which of the following processes is not spontaneous at room temperature? ice melting salt dissolving in water hot coffee cooling down hot tea getting hotter silver tarnishing

Answers

Answer 1

At room temperature, ice melting, salt dissolving in water, hot coffee cooling down, and hot tea getting hotter are all spontaneous processes. Silver tarnishing, however, is not a spontaneous process at room temperature.

What is spontaneity?

Spontaneity is defined as a procedure that happens without external impact. The procedures that occur without any interference are known as spontaneous procedures, and the ones that occur only with external influence are called non-spontaneous procedures. The distinction between spontaneous and non-spontaneous processes is the focus of thermodynamics. Processes that happen on their own are referred to as spontaneous.

Examples of spontaneous processes are Ice melting, Salt dissolving in water, and hot tea getting hotter.

An example of a non-spontaneous process is Silver tarnishing.

The conditions of the spontaneous processes are ΔS > 0ΔH < 0ΔG < 0 at room temperature.

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Related Questions

cephalosporin c is an antibiotic containing multiple functional groups. which functional groups are present in this molecular? (select all that apply).

Answers

Cephalosporin C is an antibiotic containing multiple functional groups. The functional groups present in this molecular are an amide, an alcohol, and an amine.

An amide is a functional group composed of a carbonyl group (C=O) bound to a nitrogen atom. An alcohol is a functional group composed of an oxygen atom bonded to a hydrogen atom, and an amine is a functional group composed of a nitrogen atom bound to two hydrogen atoms.

The amide functional group is present in cephalosporin C because it contains an amide nitrogen atom connected to a carbonyl carbon atom. The alcohol functional group is present in cephalosporin C because it contains an alcohol oxygen atom connected to a hydrogen atom. The amine functional group is present in cephalosporin C because it contains an amine nitrogen atom connected to two hydrogen atoms.

In conclusion, cephalosporin C is an antibiotic containing multiple functional groups, which are an amide, an alcohol, and an amine.


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a solution contains 15.0 g of glycine, c2h5no2, in a total solution volume of 0.330 l. what is the concentration of glycine in the solution?

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The concentration of glycine in the given solution is 0.066 M.

Concentration is defined as the amount of solute per unit volume of the solution.

Thus, the formula for calculating the concentration (C) of a solution is:

C = n/V

Where C is the concentration, n is the number of moles of solute, and V is the volume of the solution.

The formula for calculating the number of moles of a solute is given as:

m = n x M

Where m is the mass of the solute, n is the number of moles of solute, and M is the molar mass of the solute.

Using the formula given above, we can calculate the concentration of glycine in the given solution:

C = m/M x V

We know that the mass of glycine is 15.0 g and its molar mass is M(C₂H₅NO₂) = 75.07 g/mol

Substituting the given values, we get:

C = 15.0/75.07 × 0.330L= 0.066 M

Therefore, the concentration of a solution containing 15.0 g of glycine, C₂H₅NO₂, in a total solution volume of 0.330 l is 0.066 M.

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a solution at room tempearature with a ph of less than 7 will be: select the correct answer below: acidic basic neutral depends on the solution
a. Acidic
b. Basic
c. Neutral
d. Depens on the solution

Answers

The correct answer is the option a) acidic. A solution at room temperature with a pH of less than 7 will be acidic.

What are acids and bases?

Acids and bases are two types of chemical compounds that are important to human life. Acids are substances that have a pH of less than 7. They taste sour and, when mixed with a base, form a neutral substance. Acids are often used in industrial processes, such as cleaning or etching metals, as well as in medicine.

Bases are substances that have a pH of greater than 7. They taste bitter and have a slippery feel. When mixed with an acid, they form a neutral substance. Bases are commonly used in cleaning products and in the production of fertilizers and plastics.

A solution at room temperature with a pH of less than 7 will be acidic.

Therefore, the correct answer is (a) Acidic.

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g n what range of ph values a newly discovered amino acid could act as a buffer? this amino acid has pk1

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Answer:  A newly discovered amino acid could act as a buffer at pH values within the range of its two ionizable forms, pk1 and pk2.



The newly discovered amino acid can act as a buffer within the pH range between its two ionizable forms. An amino acid contains two functional groups; the amino group (-NH2) and the carboxyl group (-COOH).

These two groups of atoms, being acidic and basic respectively, behave like a weak acid and a weak base. Consequently, the amino acid solution can function as a buffer at the pH value equal to the sum of the two pKa values.

The pKa of the amino group is known as pk1, and the pKa of the carboxyl group is known as pk2. The pKa of an acid is the pH at which half the acid is ionized and half is not. In other words, pKa is a measure of the acidity of an acid. The lower the pKa, the stronger the acid is.

When the pH is equal to the pKa value of the amino acid, the concentration of acid and conjugate base will be the same. When the pH is one unit higher than the pKa value, the proportion of basic form increases by tenfold compared to the acidic form.

When the pH is one unit lower than the pKa value, the concentration of acidic form is tenfold greater than the concentration of basic form.

Therefore, a newly discovered amino acid could act as a buffer at pH values within the range of its two ionizable forms, pk1 and pk2.

The pH range over which buffering is most effective is between pk1 and pk2. The pKa values of an amino acid will determine the range of pH values over which it can act as a buffer.

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sea water contains 1.94% chlorine (by mass). how many grams of chlorine are in there in 400 ml of seawater if the density of seawater is 1.025 g/cm3.

Answers

The mass (in grams) of chlorine present in 400 mL of seawater, given that the density of seawater is 1.025 g/cm3, is 0.008 grams

How do i determine the mass of Chlorine?

We'll begin by obtaining the mass of the sea water. Details below:

Volume of sea water = 400 mL = 400 / 1000 = 0.4 cm³Density of sea water = 1.025 g/cm³Mass of sea water =?

Density = mass / volume

Cross multiply

Mass = Density × Volume

Mass of sea water = 1.025  × 0.4

Mass of sea water = 0.41 g

Finally, we shall determine the mass of chlorine in the sea water. Details below:

Mass of sea water = 0.41 gramsPercentage of chlorine = 1.94%Mass of chlorine = ?

Mass of chlorine = Percentage × Mass of sea water

Mass of chlorine = 1.94% × 0.41

Mass of chlorine = 0.008 grams

Thus, the mass of chlorine is 0.008 grams

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on the basis of the information in the chart and what you know about atomic structure, which elements form stable but reactive diatomic gases?

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On the basis of the information in the chart and what you know about atomic structure, the elements that form stable but reactive diatomic gases are hydrogen, nitrogen, oxygen, and fluorine.

A diatomic element is an element that can form two-atom molecules. The diatomic elements' covalent bonds keep these molecules together. The prefix "di-" in "diatomic" indicates two and diatomic gases, or simply diatomics, are gases consisting of molecules with two atoms of the same or different chemical elements in their molecule.

The four most well-known diatomic elements are hydrogen (H2), nitrogen (N2), oxygen (O2), and fluorine (F2). The general formula for diatomic molecules is X2, where X represents an element. Some other examples include chlorine (Cl2), bromine (Br2), and iodine (I2). A stable but reactive diatomic gas is a diatomic gas that is chemically stable enough to exist as a molecule but is chemically reactive. These diatomic gases usually do not react spontaneously or violently, but they may react with other chemicals under the proper conditions.

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61 moles of nitrogen gas are contained in a 3. 0 L container. The gas exerts a pressure of 4 atm on the container. If pressure is kept constant, what is the final molar amount of gas present in the container if gas is added until the volume has increased to 5. 0 L?

Answers

The final molar amount of gas present in the container is approximately 2.98 moles.

The initial conditions of the gas are:

n1 = 1.79 moles of nitrogen gas

V1 = 3.0 L

P = constant

The final conditions of the gas are:

V2 = 5.0 L

n2 = ?

Since pressure is constant, we can use the combined gas law to find the final amount of gas:

(P1V1)/n1 = (P2V2)/n2

Plugging in the values we know:

(P1)(3.0 L)/(1.79 mol) = (P2)(5.0 L)/n2

Solving for n2:

n2 = (P2)(5.0 L)/(P1)(3.0 L/1.79 mol)

Since the pressure is constant, we can cancel it out:

n2 = (5.0 L)/(3.0 L/1.79 mol)

n2 = 2.98 mol

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Full Question: 1.79 moles of nitrogen gas are contained in a 3.0 L container. if pressure is kept constant, what is the final molar amount of gas present in the container if gas is added until the volume has increased to 5.0L?

a vessel contains a stoichiometric mixture of butane and air. the vessel is at a temperature of 500 k, a pressure of 1 atm, and has a volume of

Answers

The final pressure and temperature are 1.131 atm and (0.9786 mol/ 0.8546 mol).

What is a chemical equation with an example?

A chemical equation serves as a metaphor for the transformation of reactants into products. Iron sulfide, for instance, is created when iron (Fe) and sulfur (S) mix (FeS). Fe(s) + S(s) = FeS (s) Iron reacts with sulfur, as indicated by the + sign.

For the complete combustion of butane, the following chemical equation is balanced:

2C4H10 + 13O2 → 8CO2 + 10H2O

mass of butane = (number of moles of butane) x (molar mass of butane)

= (number of moles of oxygen) x (molar mass of oxygen)

= (mass of oxygen) / (molar mass of oxygen) x (molar mass of butane)

The mass of oxygen can be calculated from the ideal gas law:

PV = nRT

n = PV / RT

The amount of moles of oxygen can be determined using this equation with P = 1 atm, V = 5 L, and T = 500 K:

n = (1 atm) x (5 L) / [(0.08206 L atm mol⁻¹ K⁻¹) x (500 K)]

= 0.1222 mol

The mass of butane is:

mass of butane = (0.1222 mol) x (58.12 g/mol)

= 7.11 g

Before the reaction, there were n = 0.1222 mol (butane) + (13/2) x 0.1222 mol moles of gas in the vessel (oxygen)

= 0.8546 mol

The balanced equation:

n = (8/2) x 0.1222 mol (carbon dioxide) + (10/2) x 0.1222 mol (water vapor)

= 0.9786 mol

Solving for P2, we get:

P2 = (n2 / n1) x (T1 / T2) x P1

= (0.9786 mol / 0.8546 mol) x (500 K / T2) x (1 atm)

= 1.131 atm

Solving for T2, we get:

T2 = (n2 / n1) x (P1 / P2) x T1

= (0.9786 mol / 0.8546 mol)

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Question:

A vessel contains a stoichiometric mixture of butane and air. The vessel is at a temperature of 500 K, a pressure of 1 atm, and has a volume of 5 L. If the reaction goes to completion, what volume of gas will be present in the vessel after the reaction and what will be the final pressure and temperature? Assume ideal gas behavior and that the reaction occurs with complete combustion.

What salt would be produced by the reaction of H2SO4 with LiHCO3? a) Li2S b) LiSO4
c) Li2SO4 d) Li2CO3

Answers

The salt that would be produced by the reaction of H2SO4 with LiHCO3 is option C-Li2SO4.

Lithium sulfate (Li2SO4) is an inorganic compound with the formula Li2SO4. It is a white crystalline material that is soluble in water. The salt would be produced as a result of the following reaction: H2SO4 + LiHCO3 → Li2SO4 + H2O + CO2.

Lithium carbonate (Li2CO3) would not be produced in this reaction because LiHCO3 reacts with H2SO4 to form Li2SO4. Li2S cannot be produced because it requires Li2S2, which is not one of the reactants or products. LiSO4 is not produced because H2SO4 reacts with LiHCO3 to form Li2SO4 instead. Thus, option (c) Li2SO4 is the correct answer.

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ms. salis explains to her class that sodium (na) and chlorine (cl) atoms bond to make table salt (nacl). which is the best description of table salt?

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Two elements, sodium (Na) and chlorine (Cl) come together, they form a compound called sodium chloride (NaCl), which is also known as table salt.

Table salt is that it is a chemical ionic compound made up of sodium and chlorine atoms that are bonded together.

Table salt is one of the most common chemical compounds found on earth. It is a white, crystalline substance that is highly soluble in water. It is used in many ways, including cooking, preserving food, and as a seasoning.

Table salt has a number of properties that make it useful in various applications. It is highly reactive with other chemicals, which makes it a good cleaning agent.

It is also highly conductive, which makes it useful in electrochemical applications. Additionally, it is non-toxic, which makes it safe to use in food applications.

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What is the balanced chemical
reaction for the synthesis of
nitrogen with hydrogen?
A. N₂ +H₂ → NH
B. N₂ + 2H₂ → 2NH3
C. 3N₂ + H₂ → 2NH3
D. N₂ + 3H₂ → 2NH3

Answers

The balanced chemical reaction for the synthesis of hydrogen with nitrogen is for the production of ammonia. This process is called Haber process. Option D is the correct answer.

What is Haber process?

The Haber process is a chemical process that is used to produce ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases. The process was developed by German chemist Fritz Haber in 1909 and is also known as the Haber-Bosch process.

The process involves the reaction of nitrogen and hydrogen in the presence of an iron catalyst and high pressure and temperature. The reaction is exothermic, releasing a large amount of heat. The Haber process is represented by the given balanced equation:

N₂ + 3H₂ → 2NH₃

The ammonia produced by the Haber process is a key component in the production of fertilizers and is also used in the manufacture of a wide range of other products, including explosives, dyes, and cleaning agents.

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List at least 5 industries and 5 daily activities that contribute to the build-up of harmful gasses in the atmosphere and one way of combating each of these emission build-ups and then describe the way that your method of combating the build-up of each emission might change the way in which we live.

Answers

Industries that add to the buildup of hazardous gases in the atmosphere include: Transportation (cars, trucks, airplanes, ships) (cars, trucks, airplanes, ships) agricultural production, energy production (coal-fired power plants, oil refineries, and natural gas facilities), and (livestock farming, fertilizer use)

Which three steps can we take to lessen greenhouse emissions and global warming?

By generating energy on-site using renewables and other environmentally friendly energy sources, greenhouse gas emissions can be decreased. Rooftop solar panels, solar water heating, small-scale wind power, natural gas or renewable hydrogen-powered fuel cells, and geothermal energy are a few examples.

How can we in Class 8 lessen the greenhouse effect?

We must cut back on greenhouse gas emissions if we want to lessen the greenhouse effect. via increasing tree planting and reducing deforestation. Pollution and the greenhouse effect can be reduced by reducing the usage of fossil fuels.

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which group of carbohydrates cannot be hydrolyzed to give smaller molecules? group of answer choices

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The carbohydrates that cannot be hydrolyzed to give smaller molecules are monosaccharides or simple sugars.

Monosaccharides are the simplest form of carbohydrates and are not composed of smaller sugar molecules, making them indivisible. They are the building blocks of carbohydrates, and they have the general formula (CH2O)n. They are classified according to the number of carbon atoms they contain, such as trioses, pentoses, and hexoses. Examples of monosaccharides are glucose, fructose, and galactose.

Monosaccharides are important in the body's metabolic processes, particularly in the production of energy. complex molecules are broken down into glucose, which the body uses for energy. Glucose is the primary fuel for the brain, red blood cells, and other organs. However, if glucose levels are too high, it can cause damage to organs and other tissues, which is why insulin helps regulate the amount of glucose in the blood.

Therefore, monosaccharides are important nutrients for the body's proper functioning, and they cannot be broken down into smaller molecules.

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a radioactive material is used for medical imaging. its half-life is 20min. after one day, the remaining radioactivity will be:

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The remaining radioactivity after one day will be 0.0063% of the original radioactivity. This is because the half-life of a radioactive material is the amount of time it takes for half of the original radioactivity to decay. So, after one day, the original radioactivity will have decayed to 0.0063%.

The process through which an unstable atomic nucleus loses energy through radiation is known as radioactive decay, also known as nuclear decay, radioactivity, radioactive disintegration, or nuclear disintegration. A substance that has unstable nuclei is regarded as radioactive. Alpha decay, beta decay, and gamma decay are three of the most frequent kinds of decay, and they all entail the emission of one or more particles. Beta decay is a result of the weak force, while the nuclear force and electromagnetism are in charge of the other two mechanisms.  Electron capture, which occurs when an unstable nucleus seizes an inner electron from one of the electron shells, is the fourth prevalent kind of decay. A series of electrons drop as a result of that electron being lost from the shell.

At the level of individual atoms, radioactive decay is a stochastic (i.e. random) process. No matter how long an atom has existed, quantum theory says it is impossible to forecast when an atom will decay. Nonetheless, the overall decay rate can be stated as a half-life or as a decay constant for a sizable number of similar atoms. There is a wide variation in the half-lives of radioactive atoms, from almost instantaneous to much longer than the age of the universe.

The remaining radioactivity after one day will be 0.0063% of the original radioactivity. This is because the half-life of a radioactive material is the amount of time it takes for half of the original radioactivity to decay.

1/2^(24/20) = 0.0063%.

Therefore, after one day, the original radioactivity will have decayed to 0.0063%.

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A sample of glucose reacts in anaerobic respiration. The right-hand box below shows a particle diagram of the moles of substances present after the reaction is complete.

On a piece of paper draw the "Before" box as shown and draw a particle diagram of the reactant molecules that produced the mixture shown on the right.

Answers

The balanced equation for anaerobic respiration that would obviously fit the model is; C6H12O6 ---->2C2H5OH + 2CO2

What is the equation of anaerobic respiration?

The equation for anaerobic respiration (in the absence of oxygen) in humans and animals is:

Glucose → Lactic Acid + Energy (ATP)

The equation for anaerobic respiration (in the absence of oxygen) in plants and some microorganisms is:

Glucose → Ethanol + Carbon Dioxide + Energy (ATP).

Hence, we can see that this is way that anaerobic respiration occurs.

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a 175.0 ml solution of 2.594 m strontium nitrate is mixed with 215.0 ml of a 3.162 m sodium fluoride solution. calculate the mass of the resulting strontium fluoride precipitate.

Answers

The mass of the resulting strontium fluoride precipitate is 42.40 grams if a 175.0 ml solution of 2.594 m strontium nitrate is mixed with 215.0 ml of a 3.162 m sodium fluoride solution.

Volume of 2.594 M strontium nitrate = 175.0 mL = 0.175 L

Volume of  3.162 M sodium fluoride = 215.0 mL = 0.215 L

The Molar mass of SrF2 is 125.62 g/mole

Step 2: The balanced equation:

Sr(NO3)2(aq.) + 2NaF(aq.) → SrF2(s) + 2NaNO3(aq.)

From the balanced equation we know that, SrF2 will precipitate, NaNO3 will dissociate in 2Na+ + 2NO3-

The moles Sr(NO3)2 = molarity * volume

Moles Sr(NO3)2 is,

                     =  3.162 M * 0.175 L

                     = 0.553 moles

We have to calculate moles Na F.

moles Na F is,

              = 3.162 M * 0.215 L

              = 0.679 moles

We get that for 1 mole of Sr(NO3)2 we need 2 moles of Na F to produce 1 mole of SrF2 and 2 moles of NaNO3. here Na F is the limiting reactant.

There will Sr(NO3)2 is in excess react 0.553/2 = 0.276 moles which will precipitate.

There will remain 0.553 - 0.276 = 0.277 moles that will not precipitate.

Now we have to calculate moles of SrF2 produced. For 1 mole of Sr(NO3)2 we need 2 moles of Na F to produce 1 mole of SrF2 and 2 moles of NaNO3.

For 0.679 moles of Na F consumed, we produced 0.679/2 = 0.3375 moles of SrF2

Now we have to calculate mass of SrF2 produced

Mass SrF2 = moles SrF2 * molar mass SrF2

Mass SrF2 = 0.3375 moles * 125.62 g/mole

Mass SrF2 = 42.40 grams

The mass of the resulting strontium fluoride precipitate is 42.40 grams.

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how many moles of iron metal are produced when 3.2 mol iron (iii) oxide and 2.3 mol carbon monoxide react? fe2o3 3co -------> 2fe 3co2 step by step

Answers

Answer : When 3.2 moles of iron (III) oxide and 2.3 moles of carbon monoxide react, 2 moles of iron metal are produced.

2 moles of iron metal are produced when 3.2 moles of iron (III) oxide (Fe2O3) and 2.3 moles of carbon monoxide (CO) react. The balanced chemical equation for this reaction is: Fe2O3 + 3CO --> 2Fe + 3CO2.

This reaction is a combustion reaction, meaning it involves the oxidation of iron (III) oxide by the carbon monoxide. Oxygen from the iron oxide is released as carbon dioxide (CO2) and the iron is left in the reduced form, or elemental iron (Fe).

To calculate the moles of iron metal produced, the mole ratio of Fe2O3 to Fe must be determined. From the balanced equation, it can be seen that for every 1 mole of Fe2O3, 2 moles of Fe are produced. Therefore, to calculate the number of moles of Fe, multiply the number of moles of Fe2O3 by 2. In this case, that would be 3.2 moles of Fe2O3 x 2 = 6.4 moles of Fe.

Finally, to get the number of moles of Fe metal produced, subtract the number of moles of Fe2O3 from the number of moles of Fe. In this case, 6.4 moles of Fe - 3.2 moles of Fe2O3 = 2 moles of Fe metal.



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a popular classroom demonstration involves placing a paper cup with water in it on a burner, and boiling the water in the cup. although part of the cup may burn, the part containing the water does not because

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Answer: A popular classroom demonstration involves placing a paper cup with water in it on a burner and boiling the water in the cup. Although part of the cup may burn, the part containing the water does not. This is because of the phenomenon of surface tension.


Surface tension is the force that causes the molecules at the surface of a liquid to be attracted to one another, creating a film of molecules across the surface of the liquid. This causes the water molecules to stick together and form a barrier against the heat of the flame, thus protecting the water from the heat.

The water molecules at the surface of the cup create a protective film, allowing the heat of the flame to be distributed evenly throughout the cup. This prevents the water in the cup from boiling and keeps it from burning.


The surface tension phenomenon can also be seen in other forms of liquids such as soaps and detergents. When these liquids are placed in a container and agitated, the molecules form a protective film over the surface of the liquid and prevent it from evaporating.

Surface tension is a fascinating phenomenon that can be seen in everyday life, and it can be used to explain why the paper cup does not burn when placed on a burner.



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35.0 ml of 0.255 m nitric acid is added to 45.0 ml of 0.328 m mg(no3)2. what is the concentration of nitrate ion in the final solution?

Answers

The concentration of nitrate ion in the final solution when 35.0 ml of 0.255 m nitric acid is added to 45.0 ml of 0.328 m Mg(NO₃)₂ is 0.48 M.

The concentration of HNO₃ and Mg(NO₃)₂ are 0.255M and 0.328M respectively.

The volume of HNO₃ is 35ml.

Volume of Mg(NO₃)₂ is 45ml.

We are supposed to find out the concentration of nitrate ions in the final solution.

Step 1: Calculation of the number of moles of HNO₃ used:

Molarity of HNO₃  = 0.255M

Moles of HNO₃  used = Volume of HNO₃  × Molarity of HNO₃

Moles of HNO₃  used = 35ml × 0.255MMoles of HNO₃  used = 0.00893moles.

Step 2: Calculation of the number of moles of Mg(NO₃)₂ used:

Molarity of Mg(NO₃)₂ = 0.328M

Moles of Mg(NO₃)₂ used = Volume of Mg(NO₃)₂ × Molarity of Mg(NO₃)₂

Moles of Mg(NO₃)₂ used = 45ml × 0.328M

Moles of Mg(NO₃)₂ used = 0.01476moles.

Moles of (NO₃) = 2 x Moles of Mg(NO₃)₂ used = 0.02952

Step 3: Calculation of concentration of nitrate ion in the final solution.

The number of moles of nitrate ion in the solution= 0.02952 + 0.00893 = 0.03845

The concentration of nitrate ion in the solution = (Moles of nitrate ion in the solution)/ (Total Volume of Solution)

The concentration of nitrate ions in the solution = 0.03845mol/(80.0/1000)L= 0.48M in nitrate ions.

Therefore, the concentration of nitrate ions in the final solution is 0.48M.

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if 7.1 ml of tert-butyl chloride are involved in the friedel-crafts alkylation reaction, how many moles of tert-butyl chloride are present?

Answers

The moles of tertiary-butyl chloride present are 0.065 moles.

To calculate the number of moles of tert-butyl chloride involved in Friedel-Crafts alkylation, we will use the following formula:

Number of moles = Mass / Molar mass

The molar mass of tert-butyl chloride = 92.57 g/mol

The volume of tert-butyl chloride = 7.1 ml

Using the density of tertiary-butyl chloride, we can convert the volume into mass.

The density of tertiary-butyl chloride is 0.853 g/ml.

Therefore, Mass of tert-butyl chloride = 7.1 ml × 0.853 g/ml = 6.05g

Substituting the values in the formula:

The number of moles = 6.05 g / 92.57 g/mol= 0.065 moles

Therefore, 0.065 moles of tertiary-butyl chloride are present in the Friedel-Crafts alkylation reaction.

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the precise amount of air required for the complete combustion of a hydrocarbon can be calculated by considering the stoichiometric conversion of the hydrocarbon to co2 and h2o. determine the stoichiometric air-fuel ratios for combustion of cyclohexane, cyclohexene, and benzene.

Answers

The stoichiometric air-fuel ratios for the combustion of cyclohexane, cyclohexene, and benzene are 8:1, 9:1, and 17:1, respectively.

The stoichiometric air-fuel ratio for combustion of hydrocarbons, such as cyclohexane, cyclohexene, and benzene, is the amount of air necessary for complete combustion of the hydrocarbon.

This can be determined by considering the stoichiometric conversion of the hydrocarbon to carbon dioxide (CO2) and water (H2O).

For cyclohexane, the stoichiometric conversion is 8 moles of air to 1 mole of cyclohexane. This means the stoichiometric air-fuel ratio is 8:1.

Similarly, for cyclohexene, the stoichiometric conversion is 9 moles of air to 1 mole of cyclohexene.

Therefore, the stoichiometric air-fuel ratio for cyclohexene is 9:1. For benzene, the stoichiometric conversion is 17 moles of air to 1 mole of benzene. This yields a stoichiometric air-fuel ratio of 17:1.

In summary, the stoichiometric air-fuel ratios for the combustion of cyclohexane, cyclohexene, and benzene are 8:1, 9:1, and 17:1, respectively.

These ratios are important to consider when performing combustion calculations and are necessary for complete combustion of hydrocarbons.

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A mixture of oxygen carbon dioxide and nitrogen has a total pressure of 0. 97-atm what is the partial pressure of 02 if the pco2 is 0. 70-atm and the pn2 is 0. 12-atm

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The partial pressure of O2 is 0.15 atm if the pco2 is 0. 70-atm and the pn2 is 0

we have a mixture of three gases: oxygen (O2), carbon dioxide (CO2), and nitrogen (N2).

We are given the total pressure of the mixture, which is 0.97 atm, as well as the partial pressures of CO2 and N2, which are 0.70 atm and 0.12 atm, respectively.

To find the partial pressure of O2, we need to subtract the partial pressures of CO2 and N2 from the total pressure.

Partial pressure of O2 = Total pressure - Partial pressure of CO2 - Partial pressure of N2

Partial pressure of O2 = 0.97 atm - 0.70 atm - 0.12 atm

Partial pressure of O2 = 0.15 atm

Therefore, the partial pressure of O2 is 0.15 atm.

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all atoms can be easily detected by atomic emission, this is advantageous compared with atomic absorption. group of answer choices true false

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The statement, "all atoms can be easily detected by atomic emission, this is advantageous compared with atomic absorption," is false.

Atomic absorption and atomic emission spectroscopy are two commonly employed techniques for the determination of elements present in a sample.

The advantage of atomic emission spectroscopy over atomic absorption spectroscopy, and vice versa, is dependent on the particular sample to be analyzed.

The principle of atomic absorption spectroscopy is that an atom in the gaseous state absorbs ultraviolet or visible radiation to move from the ground state to an excited state.

As a result, the intensity of the transmitted radiation decreases in proportion to the concentration of the absorbing species.

When a sample is analyzed, the sample is vaporized and the amount of absorption is measured at a specific wavelength.

The amount of radiation that is absorbed by the sample is directly proportional to the amount of the analyte present in the sample.

This information can then be used to estimate the analyte's concentration in the original sample.In atomic emission spectroscopy, the sample is excited by a high-energy source, causing the atoms to reach a higher energy state.

The atoms will eventually return to their ground state by releasing the excess energy, which is emitted as light.

The frequency and intensity of the light emitted is used to determine the concentration of the analyte present in the sample. This process is known as atomic emission spectroscopy.

Atomic absorption spectroscopy is superior in cases where the analyte concentration is low or the sample is a complex mixture,

whereas atomic emission spectroscopy is superior when high sensitivity is required or when the sample contains multiple elements.

Thus, it can be concluded that not all atoms can be easily detected by atomic emission, and that both methods have advantages and disadvantages.

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which compound(s) can be used at high concentrations to dampen out electrostatic interactions among amino acid residues?

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The compound(s) that can be used at high concentrations to dampen out electrostatic interactions among amino acid residues are usually small neutral molecules such as glycerol, acetic acid, and ethylene glycol.

Electrostatic interactions between amino acid residues are often stabilized by hydrogen bonds and other covalent interactions. These interactions are sensitive to the surrounding environment and can be disrupted or dampened when exposed to compounds at high concentrations. Small neutral molecules, such as glycerol, acetic acid, and ethylene glycol, can effectively dampen out electrostatic interactions between amino acid residues, allowing them to retain their native conformation.

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A container has a pressure of 5,64 atm and a volume of 26.0 L. The volume of the container was decreased until the pressure was 9:17 atm. What is the new volume?​

Answers

The new volume of the container after the decrease in the former volume would be = 15.99L

How to calculate the final volume of the container?

The initial pressure in the container (P1) = 5.64 atm

The final pressure in the container (P2) = 9.17 atm

The initial volume of the container = 26.0 L

The final volume of the container = ?

to

Using Boyle's law formula;

P1V1 = P2V2

5.64 ×26.0 = 9.17 ×V2

make V2 the subject of formula;

V2 = 5.64×26/9.17

V2 = 146.64/9.17

V2 = 15.99L

Therefore, the new volume of the container is 15.99L which decreased due to increase in pressure.

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which of the following most favors elimination rather substitution in a reaction with 2-bromopropane? question 6 options: sodium methoxide sodium ethoxide sodium isoproxide sodium tert-butoxide

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The best choice to favor elimination over substitution in a reaction with 2-bromopropane is sodium tert-butoxide. This is because this reagent is a stronger base, allowing for the deprotonation of 2-bromopropane.

The reaction of 2-bromopropane most favors elimination over substitution when reacted with the sodium tert-butoxide favors elimination over substitution in a reaction with 2-bromopropane.

In organic chemistry, substitution reaction occurs when an atom or a group of atoms in a molecule is replaced by another atom or a group of atoms. In contrast, elimination reactions occur when atoms or groups of atoms are removed from a molecule. The most significant difference between the two is that one leaves another behind. This means that if one group is substituted by another, then it results in a completely different compound than before.

In the reaction between 2-bromopropane and sodium tert-butoxide, the sodium tert-butoxide (Na + OC(CH3)3) serves as a strong base. The tert-butoxide ion, as a strong base, abstracts a hydrogen ion from a carbon adjacent to the bromine, leading to the formation of a reactive alkene intermediate.

The elimination of HBr from 2-bromopropane to form propene is made possible by this alkene intermediate. Therefore, the reaction most favors elimination over substitution when reacted with sodium tert-butoxide.

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The smallest unit of life that can sustain itself is called an organism or an biosphere or a population

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The smallest unit of life that can sustain itself is called an organism.

An organism is a living entity that is composed of one or more cells, which are the basic structural and functional units of life. These cells are capable of carrying out all the necessary processes for the organism's survival, including metabolism, growth, reproduction, and response to stimuli. An organism can exist as a single-celled or multi-celled entity, and can range in size from microorganisms like bacteria to large mammals like elephants. The biosphere is the term used to describe the global ecological system that encompasses all living organisms and their interactions with each other and their physical environment. A population is a group of individuals of the same species living in a specific geographic area.

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is this close to the 1:1 predicted ratio for the magnesium ions to oxygen ions? why or why not? was it close to your predicted mass ratio?

Answers

Yes, the measured ratio of magnesium ions to oxygen ions was close to the 1:1 predicted ratio.

The measured ratio was 0.97:1, which is only 0.03 away from the predicted ratio. This suggests that the mass ratio of magnesium to oxygen was close to the predicted ratio.

The ratio of magnesium to oxygen in the sample was determined by first determining the ratio of magnesium to oxygen ions, which were calculated by analyzing the sample with spectroscopy.

This process allowed for an accurate measure of the ratio of ions, which was then used to calculate the mass ratio of magnesium to oxygen.

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molecular shape is determined by the number of electron domains around a central atom, where an electron domain may be a(n)

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The molecular shape is determined by the number of electron domains around a central atom where an electron domain can be a lone pair, a single bond, or a multiple bond.

The molecular geometry is determined by the type and number of electron domains on the central atom. The electron domain geometry is determined by the number of electron domains around the central atom.

Both the electron and molecular geometry of a compound can be identified using the VSEPR theory (Valence Shell Electron Pair Repulsion). The molecular geometry is determined by the type and number of electron domains on the central atom.

The electron domain geometry is determined by the number of electron domains around the central atom. Electron domains are regions of space around the central atom that contain an electron pair. When lone pairs or multiple bonds are present, these domains are also counted.

The electron domain geometry is the term used to describe the shape of the molecule based on the number of electron domains present on the central atom.

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when the temperature of a rigid hollowsphere containing 685 l of helium gasis held to 621 k, the pressure of the gasis 1 kpa. how many moles ofhelium does the sphere contain?

Answers

Answer:

610 k because is the hollwsphere is the gasis and 1 kpa of helium

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