The best description of a star is: "a celestial body that emits light due to nuclear reactions and is composed of gases."
Stars are massive, luminous spheres of plasma held together by their own gravity. They are not composed of rock and metal with a high density, nor are they too big to be an asteroid but too small to be a planet. Stars do rotate on an axis and revolve around a central point, but their defining characteristic is their ability to emit light and heat through the process of nuclear fusion.
What is celestial body?
A celestial body is any object that exists in space, such as a planet, moon, asteroid, comet, star, or galaxy. Celestial bodies are natural objects that are not made by humans, and they are typically studied by astronomers and other scientists who are interested in learning more about the structure, composition, and behavior of the universe. Celestial bodies can be located within our solar system or in distant regions of space, and they can range in size from tiny rocks to massive stars and superclusters of galaxies.
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Complete question is: The best description of a star is: "a celestial body that emits light due to nuclear reactions and is composed of gases."
How many grams of chlorine gas can be liberated from the decomposition of 169. 0 g. Of AuCl3
169.0 g of [tex]AuCl _{3}[/tex] can liberate 118.4 g of [tex]Cl_{2}[/tex] gas upon decomposition. The molar mass of [tex]AuCl _{3}[/tex] is 303.33 g/mol, which means that 1 mole of [tex]AuCl _{3}[/tex]contains 3 moles of chlorine (3 atoms of chlorine).
To determine the moles of [tex]AuCl _{3}[/tex]in 169.0 g, we divide the mass by the molar mass:
169.0 g / 303.33 g/mol = 0.557 moles of [tex]AuCl _{3}[/tex]
Since each mole of [tex]AuCl _{3}[/tex] produces 3 moles of chlorine, the total moles of chlorine that can be liberated from the decomposition of 0.557 moles of [tex]AuCl _{3}[/tex]is:
0.557 moles x 3 = 1.671 moles of [tex]Cl_{2}[/tex]
Finally, we use the molar mass of chlorine ([tex]Cl_{2}[/tex]), which is 70.90 g/mol, to convert the moles of [tex]Cl_{2}[/tex]to grams:
1.671 moles x 70.90 g/mol = 118.4 g of [tex]Cl_{2}[/tex]
Therefore, 169.0 g of [tex]AuCl _{3}[/tex]can liberate 118.4 g of [tex]Cl_{2}[/tex]gas upon decomposition.
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At standard pressure, which substance becomes less soluble in water as temperature increases from 10.°C to 80.°C?
a. KCl
b. HCl
c. NaCl
d. NH4Cl
At standard pressure, which substance becomes less soluble in water as temperature increases from 10.°C to 80.°C is : d. NH4Cl
What happens to solubility with temperature?Solubility of substance increases with temperature, as higher temperatures allow more particles to dissolve in solvent. However, there are some exceptions, where solubility decreases with increasing temperature.
In this case, we are looking for substance that becomes less soluble in water as temperature increases from 10°C to 80°C at standard pressure.
The correct answer is d. NH4Cl.
At standard pressure, solubility of NH4Cl decreases with increasing temperature due to its endothermic dissolution process. As temperature increases, heat absorbed by the solution also increases, which makes dissolution process less favorable. Therefore, solubility of NH4Cl decreases with increasing temperature.
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Given 30 grams NaBr how many grams of Cl2 are required to complete this reaction?
2NaBr + Cl2 = 2NaCl + Br2
Answer:
10.3 grams
Explanation:
The balanced equation shows that 1 mole of Cl2 reacts with 2 moles of NaBr. To find out how much Cl2 is required to react with 30 grams of NaBr, we need to convert grams to moles.
First, we need to find the molar mass of NaBr:
NaBr = 23 + 79.9 = 102.9 g/mol
Now we can calculate the number of moles of NaBr:
30 g NaBr ÷ 102.9 g/mol = 0.291 moles NaBr
From the balanced equation, we know that 1 mole of Cl2 reacts with 2 moles of NaBr. Therefore, we need half as many moles of Cl2 as we have moles of NaBr:
0.291 moles NaBr ÷ 2 = 0.1455 moles of Cl2
Finally, we can convert moles of Cl2 to grams using its molar mass:
Cl2 = 35.5 x 2 = 71 g/mol
0.1455 moles Cl2 x 71 g/mol = 10.3 grams of Cl2
Therefore, 10.3 grams of Cl2 are required to react completely with 30 grams of NaBr in this reaction.
a sample of 500 oxygen atoms contains a mixture of 20o (20.0040754 amu) and 23o (23.01570 amu). the weighted average of oxygen atoms 22.76272353 amu. how many 20o atoms are present in this sample?
The number of 20o atoms present in the given sample of 500 oxygen atoms is 301.
The weighted average is a measure of central tendency that takes into account the varying weights (or importance) of the different values in a data set. A weighted average is calculated by multiplying each value by its weight and dividing it by the sum of the weights. It is also referred to as a weighted mean or weighted arithmetic mean. The formula for the weighted average is,
weighted average = (w1x1 + w2x2 + w3x3 + … wn xn) / (w1 + w2 + w3 + … wn)
Where,x1, x2, x3, … xn are the values of the individual observations w1, w2, w3, … wn are the respective weights of the individual observations to solve the given problem. A sample of 500 oxygen atoms contains a mixture of 20o (20.0040754 amu) and 23o (23.01570 amu). The weighted average of oxygen atoms is 22.76272353 amu. We have to find how many 20o atoms are present in this sample.
Here,20o atoms have a mass of 20.0040754 amu23o atoms have a mass of 23.01570 amu
The average mass of 500 oxygen atoms is 22.76272353 amu. Therefore,
500 oxygen atoms weigh= 500 x 22.76272353 = 11381.36176 amu.
average atomic mass of oxygen = 20o atoms (20.0040754 amu) and 23o atoms (23.01570 amu)
Atomic mass of oxygen = [(number of 20o atoms x mass of 20o atoms) + (number of 23o atoms x mass of 23o atoms)] / Total number of oxygen atoms
Let's consider a number of 20o atoms to be ‘x’.Therefore, the number of 23o atoms = (500 - x). Substituting the given values in the above formula, we get,
22.76272353 = [(x x 20.0040754) + ((500 - x) x 23.01570)] / 50022.76272353 = (20.0040754x + 11507.825 - 23.01570x) / 50022.76272353 x 500 = 11507.825 - 3.0116246xx = 301.05 ≈ 301Number of 20o atoms = 301.
Hence, 301 number of 20o atoms are present in this sample.
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the sodium atom loses 1 electrons when it reacts with something. the electron configuration of the sodium ion is the same as the electron configuration of
The sodium atom loses 1 electron when it reacts with something. The electron configuration of the sodium ion is the same as the electron configuration of the noble gas neon.
An electron is a negatively charged subatomic particle that orbits the nucleus of an atom.
The electrons that orbit the nucleus of an atom are arranged in shells, which are concentric circles around the nucleus, in what is known as the electron configuration. Electron configuration is the arrangement of electrons in the orbitals of an atom or molecule in its ground state.
Sodium is a chemical element with the symbol Na and atomic number 11.
Sodium is a soft, silvery-white metal that is extremely reactive.
Sodium readily loses one electron to form a positively charged ion, and it is this characteristic that makes it an important component of many compounds.
In a neutral atom, a sodium atom has eleven electrons, with the electron configuration being 1s²2s²2p⁶3s¹.
When a sodium atom loses an electron, it becomes a positively charged sodium ion with a 1+ charge.
When a sodium atom loses an electron, the electron configuration of the sodium ion is the same as that of the noble gas neon. Therefore, the electron configuration of a sodium ion is 1s²2s²2p⁶.
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how much volume in ml will you need to take from 3.8 m concentrated stock solution if you would like to prepare a diluted 0.9 solution with 100 ml? report and round your answer to a whole integer.
You will need to take 24 mL of the concentrated solution and add it to 76 mL of solvent to prepare 100 mL of a 0.9 M diluted solution.
The given data is as follows:
concentrated stock (C1)= 3.8 m
diluted solution (C2) = 0.9 m
Total available solution (V2)= 100ml
To find out the volume of the concentrated solution needed, we must use the formula
V1 x C1 = V2 x C2
By substituting the values we can get,
V1 x 3.8 M = 100 mL x 0.9 M
V1 = (100 mL x 0.9 M) / 3.8 M
V1 = 23.68 mL
Rounding to the nearest whole integer as mentioned we get:
V1 = 24 mL
Therefore, we can conclude that you will need to take 24 mL of the concentrated solution.
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explain why oxygen forms 2 bonds to hydrogen to make a water molecule, while nitrogen forms 3 bonds to make a molecule of ammonia
Oxygen and nitrogen are both nonmetals, meaning they form covalent bonds when they react.
Oxygen forms two covalent bonds with hydrogen because it has six valence electrons and needs two more electrons to complete its octet. Nitrogen has five valence electrons and needs three more electrons to complete its octet, so it forms three covalent bonds with hydrogen. The chemical formula for a water molecule is H2O, meaning that two hydrogen atoms are bonded to one oxygen atom. The chemical formula for ammonia is NH3, meaning that three hydrogen atoms are bonded to one nitrogen atom. The bond between hydrogen and oxygen is a polar covalent bond, while the bond between hydrogen and nitrogen is a non-polar covalent bond. This is due to the difference in electronegativity between oxygen and nitrogen, which causes oxygen to be more electronegative than nitrogen.
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Chemical equilibrium occurs when free energy exists in the _____.
highest possible value
lowest possible value
The statement that correctly defines chemical equilibrium is, "Chemical equilibrium occurs when free energy exists in the lowest possible value."
Chemical equilibrium is a state in which the forward and reverse chemical reactions take place at the same rate. The point at which this occurs is referred to as the equilibrium point.
The forward and backward reactions that result in chemical equilibrium continue to occur; they just occur at the same speed, resulting in no net change in the system's chemical concentration over time.
The Gibbs free energy of a chemical reaction determines the spontaneity of the reaction. If the ΔG value is positive, the reaction is non-spontaneous; if the ΔG value is negative, the reaction is spontaneous; and if the ΔG value is zero, the system is in equilibrium. In equilibrium, the free energy exists in the lowest possible value.
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a 67.0 ml aliquot of a 0.600 m stock solution must be diluted to 0.100 m. assuming the volumes are additive, how much water should be added?
To dilute a 67.0 ml aliquot of a 0.600 m stock solution to 0.100 m, 402.0 ml of water must be added.
To dilute a 67.0 ml aliquot of a 0.600 m stock solution to 0.100 m, the amount of water to be added can be calculated using the formula: M1V1 = M2V2.
M1 = 0.600 m, V1 = 67.0 ml, M2 = 0.100 m, V2 = Unknown
V2 = (M1V1) / M2
V2 = (0.600 x 67.0) / 0.100
V2 = 402.0
When a stock solution is diluted, it is mixed with a solvent such as water. The amount of solvent (in this case, water) to be added can be calculated using the above formula.
The initial volume (V1) and the concentration (M1) of the stock solution are known, while the final concentration (M2) and the final volume (V2) are unknown.
The formula can be used to calculate the amount of solvent to be added in order to reach the desired concentration.
The initial volume of the stock solution was 67.0 ml, and the initial concentration was 0.600 m. The desired concentration was 0.100 m.
When the formula was used, it was found that 402.0 ml of water must be added in order to reach the desired concentration.
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Classify the bond types for each of the following pairs of atoms (PLEASE ANSWER ALL AND EXPLAINNN :)
A.) Hydrogen and nitrogen
B.) Carbon and sulfur
C.) fluorine and fluorine
D.) beryllium and oxygen
Answer:
a.polar covalent
b.ovalent
c.covalent
d.covalent
Explanation:
a.the atomic number of nitrogen is 7 and atomic number of hydrogen is 1, so the type of bond firmed btw them is called polar covalent
b.The total valence electrons in sulphur atom are 6.thus, one atom of carbon forms two *Covalent bonds* with sulphur atoms each in order to complete it octet. Hence, the bond btw carbon and sulfur us covalent bond
c.The two fluorine atom form a stable F molecule by sharing two element ; the linkage ² is called a Covalent bonds
a 2.90 m solution of methanol (ch3oh) in water has a density of 0.984 g/ml what are the a) mass percent, b) molarity, and c) mole percent of solute in this solution?
A 2.90 m solution of methanol (ch3oh) in water has a density of 0.984 g/ml has no mass percentage, The molarity of the solution is 0.000872 M and the mole percent of the solute in the solution is 0.0018%.
a) Mass percent
The mass percent of solute in the solution is the mass of the solute divided by the mass of the solution, then multiplied by 100. The mass percent of the solute in the given solution is computed below:
Mass of the solution = Volume of the solution × Density of the solution
= 2.90 L × 0.984 g/mL= 2.8476 g
Mass of the solute = Mass of the solution - Mass of water= 2.8476 g - (2.90 L × 1000 g/L) = -5.40 g
Mass percent = (mass of solute / mass of solution) × 100
= (-5.40 g / 2.8476 g) × 100= -189.89% (not possible)
Therefore, the mass percent of solute in the solution is not possible.
b) Molarity
The number of moles of solute present in the given solution is first calculated:
Molar mass of CH3OH = 12.01 + 3(1.01) + 16.00 = 32.04 g/mol
Mass of CH3OH in solution = Volume of solution × Density of solution × Mass percent of solute / 100
= 2.90 L × 0.984 g/mL × 2.89% / 100 = 0.0810 g
Moles of CH3OH in solution = mass of CH3OH / molar mass of CH3OH
= 0.0810 g / 32.04 g/mol= 0.00253 mol
Therefore, the molarity of the solution:
Molarity = Moles of solute / Volume of solution in liters
= 0.00253 mol / 2.90 L
=0.000872 M or 8.72 x 10^-4 Mc)
Therefore, the molarity of the solution is 0.000872 M or 8.72 x 10^-4 Mc)
c) Mole percent
The mole percent of the solute in the solution is computed as follows:
Mole fraction of solute = Moles of solute / Moles of solute + Moles of solvent
= 0.00253 / (0.00253 + 139.53)
= 0.000018 mole
Mole percent of solute = (mole fraction of solute × 100)
= (0.000018) × 100= 0.0018%
Therefore, the mole percent of the solute in the solution is 0.0018%.
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How many liters does one mole of gas occupy at STP?
At STP, one mole (6.02 × [tex]10^{23}[/tex] representative particles) of any gas occupies a volume of 22.4 L . A mole of any gas occupies 22.4 L at standard temperature and pressure (0°C and 1 atm).
what must be true of a spontaneous process? a) enthalpy must increase. b) there will be a flow of heat. c) entropy of the universe increases. d) entropy of the system increases. e) entropy of the surroundings increases.
The statement that must be true of a spontaneous process is C) entropy of the universe increases as heat is generated.
In thermodynamics, a spontaneous process is one that occurs on its own without any external intervention or help. The Second Law of Thermodynamics states that entropy, a measure of disorder and randomness, must increase for a spontaneous process to occur. This means that the entropy of the universe must increase, as entropy of the system and its surroundings could both increase, decrease, or stay the same.
In terms of enthalpy, which is the energy required to bring the system to its initial state, it could increase, decrease, or stay the same, depending on the process. The flow of heat is related to entropy, as a change in entropy is accompanied by a flow of heat. This means that in a spontaneous process, there must be a flow of heat, but it doesn’t necessarily have to be from the system to the surroundings.
To summarize, for a spontaneous process to occur, entropy of the universe must increase and the flow of heat must be present, but the enthalpy of the system can remain the same, increase, or decrease, depending on the process.
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PLEASE HELP THIS IS URGENT
The equation for the production of sulfur trioxide gas from sulfur dioxide (57.50 g) and oxygen (20.0 L) using the ideal gas law indicates;
The volume of sulfur trioxide that will be formed at STP is 20.1 L
The volume of sulfur trioxide formed at 15.0°C and 98920 Pa is 21.7 L
What is the ideal gas law?The ideal gas law is an equation of state that describes an ideal gas behavior. It relates the pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) of the gas and the universal gas constant. The equation is written as P·V = n·R·T
The balanced chemical equation for the reaction is: 2SO₂ (g) + O₂ (g) --> 2SO₃ (g)
First, we need to convert the given amounts of reactants to moles. We can do this by using the molar mass of SO₂ (64.07 g/mol) and the ideal gas law for O₂ (P·V = n·R·T). At STP (Standard Temperature and Pressure), the temperature is 0°C (273.15 K) and the pressure is 1 atm (101325 Pa). The gas constant R is 8.314 J/Kmol.
The number of moles of SO₂ is: 57.50 g/(64.07 g/mol) = 0.897 moles
The number of moles of O₂ is; (101325 Pa)·(20.0 L)/(8.314 J/K.mol)·(273.15 K) = 0.892 moles
Since the ratio of SO₂ to O₂ in the balanced equation is 2:1, SO₂ is the limiting reactant and will determine the amount of product formed.
The number of moles of SO₃ produced is; (0.897 mol SO₂)·(2 mol SO₃/2 mol SO₂) = 0.897 mol (Which is based on the number of moles of SO₂ in the reactant side of the equation)
At STP, one mole of any gas occupies a volume of 22.4 L, so the volume of SO₃ produced at STP is: (0.897 mol) × (22.4 L/mol) ≈ 20.1 LTo find the volume of SO₃ at 15°C and 98920 Pa, we can use the ideal gas law again; P·V = n·R·T
V = (n·R·T)/P = ((0.897 mol)·(8.314 J/K.mol)·(288.15 K))/(98920 Pa) ≈ 21.7 LTherefore, the volume of sulfur trioxide formed at STP is 20.1 L and at 15°C and 98920 Pa is 21.7 L
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calculate the ph for each case in the titration of 50.0 ml of 0.210 m hclo(aq) with 0.210 m koh(aq). use the ionization constant for hclo.
The initial pH of the titration is 2.50 and the final pH of the titration is: -1.67.
To calculate the pH for each case in the titration of 50.0 mL of 0.210 M HClO (aq) with 0.210 M KOH (aq), you must first use the ionization constant for HClO. The ionization constant for HClO is equal to 1.5 x 10-2. Now, you can calculate the pH of the titration.
At the beginning of the titration, the pH can be determined by the initial concentration of HClO (0.210 M). Since HClO is a weak acid, it partially dissociates in water, releasing hydrogen ions. The [H+] is equal to the HClO initial concentration multiplied by the ionization constant: [tex][H+] = 0.210 x 1.5 x 10-2 = 3.15 x 10-3[/tex]
The pH can be determined by the negative logarithm of the [tex][H+], or pH = -log[H+][/tex]. So, the initial pH of the titration is [tex]-log (3.15 x 10-3) = 2.50.[/tex]
As the titration proceeds, the pH will increase due to the addition of KOH, a strong base. The final pH of the titration can be calculated in the same manner. At the equivalence point, the [H+] is equal to the KOH initial concentration multiplied by the ionization constant:[tex][H+] = 0.210 x 1 = 0.210.[/tex]
The pH of the equivalence point is [tex]-log (0.210) = -1.67.[/tex] To summarize, the initial pH of the titration is 2.50 and the final pH of the titration is -1.67.
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which is the best colored indicator to use in the titration of 0.1 m ch3cooh(aq) with naoh(aq)? why? look up ka values in the appendix. indicator pka bromocresol green 4.8 bromothymol blue 6.8 phenolphthalein 9.2 group of answer choices
Answer:
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a sample of neon has a volume of 40.81 m3 at 23.5c. at what temperature, in kelvins, would the gas occupy 50.00 cubic meters? assume pressure is constant. a. 363.27 k b. 230.54 k c. 242.0 k d. 28.79 k
At the temperatute of 363.27 K the sample of the gas Neon would occupy a volume of 50.00 cubic meters. Therefore option A can be considered correct.
Using the combined gas law in order to solve this problem
(P₁V₁)/T₁ = (P₂V₂)/T₂
( P is the pressure, V is the volume, and T is the temperature)
Since the pressure is constant, we can simplify the equation to:
V₁/T₁ = V₂/T₂
After inserting the values given in the problem equation,
V₁ = 40.81 m³
T₁ = 23.5°C + 273.15 = 296.65 K
V₂ = 50.00 m³
We can solve for T₂= (V₂/V₁) × T₁
T₂ = (50.00/40.81) × 296.65
T₂ = 363.27 K
Hnce, the temperature in kelvins at which the gas would occupy the volume of 50.00 cubic meters is calculated out to be 363.27 K.
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a polar covalent bond is associated with which of the following? group of answer choices interactions between nuclei unequal sharing of electrons equal sharing of electrons the transfer of electrons
A polar covalent bond is associated with unequal sharing of electrons.
A polar covalent bond is a covalent bond in which electrons are not equally shared between the bonded atoms. It is formed when two or more atoms share electrons in such a manner that the nucleus of one atom exerts a greater attraction on the electrons than the other atom.
As a result of the unequal sharing of electrons, the atoms have partial charges. In polar covalent bonds, the electrons spend more time near the atom with a stronger nucleus. As a result, one atom in a polar covalent bond becomes partially negative, and the other becomes partially positive. Polar covalent bonds can be found in a variety of compounds, including water, ammonia, and hydrogen chloride, among others.
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sodium metal is also readily oxidized by oxygen. if the product of the reaction were dissolved in water, what would be the color of the litmus for a litmus test? explain. what is the product?
cacl2 is a salt that forms as the result of what type of bond? multiple choice ionic nonpolar polar hydrogen covalent
CaCl2 is a salt that forms as the result of ionic bonds. An ionic bond is a bond that forms between a metal and a nonmetal when they react. One of the atoms will be electronegative, while the other will be electropositive.
When an atom is electropositive, it is more likely to give up its electrons, whereas an electronegative atom is more likely to take up an electron or electrons.
A covalent bond is formed between two nonmetal atoms when they react. Unlike an ionic bond, which occurs between a metal and a nonmetal, a covalent bond occurs between two nonmetal atoms.
The electrons are shared in a covalent bond, with each atom receiving one. As a result, both atoms have a stable number of electrons in their outermost shell.
A bond in which one atom is more electronegative than the other and thus attracts electrons more strongly is known as a polar bond.
The positive end of the molecule is the less electronegative end, and the negative end is the more electronegative end.
A hydrogen bond is a weak bond that occurs between a hydrogen atom and an electronegative atom such as nitrogen, oxygen, or fluorine.
Despite being weak, hydrogen bonds are crucial in many biological processes, such as the formation of DNA. When two atoms are identical, the bond between them is nonpolar.
In the case of a covalent bond, this occurs when the two atoms share electrons equally.
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a 250.ml sample of oxygen gas is collected over water at 25oc and 760.0 torr pressure. what is the pressure of the dry gas alone? (vapor pressure of water at 25oc is 23.8torr)
The pressure of the dry gas alone can be calculated using the ideal gas law: PV = nRT and the pressure is 736.2 torr.
The pressure of dry gas alone is 736.2 torr. Step-by-step explanation: Given that, the Volume of oxygen gas = 250 ml. Temperature = 25 oC Pressure = 760 torr, Vapor pressure of water at 25 oC = 23.8 torrTo find: The pressure of the dry gas alone.
Formula used,V2 = (P1 - P2) * (V1 - Vw) / P2Where,V2 = Volume of gas aloneP1 = Pressure of gas collectedP2 = Vapor pressure of water at temperature T1V1 = Volume of gas collected Vw = Volume of water vapor formedCalculation,P1 = 760 torrP2 = 23.8 torrV1 = 250 mlVw = V1 * P2 / P1= 250 * 23.8 / 760= 7.84 mlV2 = (P1 - P2) * (V1 - Vw) / P2= (760 - 23.8) * (250 - 7.84) / 760= 231.82 mlPressure of dry gas alone = P1 * V2 / V1= 760 * 231.82 / 250= 736.2 torr.
Hence, the pressure of the dry gas alone is 736.2 torr.
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how many grams of glucose (c6h12o6 ) would you need to prepare 1.25 l of 4% (by mass) glucose solution? assume that the percentage by mass is the number of grams per 100 ml of solution.
To prepare 1.25 L of 4% (by mass) glucose solution, the amount of glucose (C6H12O6) needed is approximately 50 grams.
Glucose is a monosaccharide with the molecular formula C6H12O6. It is also known as dextrose, grape sugar, or blood sugar. Glucose is produced by photosynthesis in green plants and is the main source of energy for the cells of the human body. Glucose is a carbohydrate with a chemical structure similar to other sugars.
A 4% (by mass) glucose solution is a solution that contains 4 grams of glucose in 100 ml of water. It is also known as a 4% weight/volume (w/v) solution. This solution is often used in medical settings to treat hypoglycemia, or low blood sugar levels.
To calculate the amount of glucose (C6H12O6) needed to prepare a 4% (by mass) glucose solution:
Step 1: Convert the volume of the solution to milliliters.1.25 L x 1000 mL/L = 1250 mL
Step 2: Calculate the mass of glucose needed to make a 4% (by mass) solution.4 g glucose/100 mL solution x 1250 mL solution = 50 g glucose
Therefore, approximately 50 grams of glucose (C6H12O6) would be needed to prepare 1.25 L of a 4% (by mass) glucose solution.
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A student investigates the number of particles of water that exist in a closed test tube throughout the phase
change of liquid to gas.
How many particles will be in the test tube after the water vaporizes and turns into a gas?
The number of particles of water that exist in a closed test tube after the water vaporizes and turns into a gas will be the same as the number of particles before the phase change.
This is because during the phase change, the molecules of water simply change their state from liquid to gas.the phase change from liquid to gas does not involve any change in the number of molecules, only a change in the physical state of the molecules. The molecules do not disappear or gain additional molecules from outside the test tube. As such, the number of particles of water in the test tube after the phase change is the same as before the phase change.
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an ion with a positive charge is called a(n) , whereas an ion with a negative charge is called a(n)
An ion with a positive charge is called a cation, whereas an ion with a negative charge is called an anion.
An ion with a positive charge is called a cation, whereas an ion with a negative charge is called an anion. Ions are atoms or molecules that have lost or gained one or more electrons, giving them a positive or negative charge. Ions play a crucial role in many chemical reactions and are found in a variety of materials.
Ions are classified as cations and anions based on their charge.
Cations: Cations are ions that have lost one or more electrons and have a positive charge. Sodium ion (Na+) is an example of a cation. Sodium atoms lose one electron, giving them a positive charge.
Anions: Anions are ions that have gained one or more electrons and have a negative charge. Chloride ion (Cl-) is an example of an anion. Chlorine atoms gain an electron, giving them a negative charge.
Thus, an ion with a positive charge is called a cation, whereas an ion with a negative charge is called an anion.
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Calculating volume (formula) and density of regular shaped objects
Please help I need to complete this assignment fast :( I’m not sure on how to do it, If you don’t know how to do it don’t answer pls
The density of the unknown sample is 1.025 g / mL and its salt composition is 3.55 %.
How to solve
PART A: Density of a regular shaped object:
Trial 1: mass of the object = 162.20 g
volume of object = L x H x W = 4.90 cm x 3.90 cm x 2.90 cm
= 55.419 cm^3
Therefore density of the object = mass / volume = 162.20 g / 55.419 cm^3
= 2.9268 g/cm^3
trial 2: mass of the object = 162.18 g
volume of object = L x H x W = 4.89 cm x 3.90 cm x 2.88 cm
= 54.92448 cm^3
Therefore density of the object = mass / volume = 162.18 g / 54.92448 cm^3
= 2.9528 g/cm^3
Average = [ 2.9268 + 2.9528 ] /2 = 5.8796 / 2 = 2.9398 g / cm^3 = 1.94 g / cm^3.
The accepted value is 2.73 g / cm^3 for aluminium. The difference is 0.21
% error = 100 x difference / accepted value = 100 x 0.21/2.73 = 7.7 %.
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Part B: Determination of density of an irregular shaped object:
Trial 1:
mass of the marble chips = 10.25 g
Volume of the marble chip = final volume of water - initial volume of water
= 53.8 - 50 = 3.8 mL
Therefore density of marble chip = mass / volume = 10.25 g / 3.8 mL
= 2.697 g / mL
Trial 2:
mass of the marble chips = 10.32 g
Volume of the marble chip = final volume of water - initial volume of water
= 53.9 - 50.1 = 3.8 mL
Therefore density of marble chip = mass / volume = 10.32 g / 3.8 mL
= 2.716 g / mL
Average = [2.697 + 2.716] / 2 = 5.413 / 2 = 2.71 g / mL
The accepted density of marble chip = 2.70 g / mL The difference is 0.01
% error = 100 x difference / accepted value = 100 x 0.01/ 2.70 = 0.37 %.
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PART C: Determination of density of saline solution:
Trial 1:
Volume of the saline solution = 10 mL
mass of the saline solution = finall mass - initial mass
= 35.66 - 25.36 = 10.3 g
Density of the saline solution = mass / volume = 10.3 g / 10 mL = 1.03 g / mL
Trial 2:
Volume of the saline solution = 10 mL
mass of the saline solution = finall mass - initial mass
= 35.55 - 25.35 = 10.2 g
Density of the saline solution = mass / volume = 10.2 g / 10 mL = 1.02 g / mL
Average =[ 1.03 + 1.02 ] / 2 = 1.025 g / mL
Thus the unknown sample B has the density of 1.025 g / mL.
The composition of salt in this solution can be determined by interpolation.
salt % = 0 + 5 x [ 1.025-0.998] / [1.036 - 0.998] ( using the values given in the table )
= 0 + 5 x 0.027 / 0.038
= 3.55 %.
Thus the density of the unknown sample is 1.025 g / mL and its salt composition is 3.55 %.
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What is the heat, q , in joules transferred by a chemical reaction to the reservoir of a calorimeter containing 155 g of dilute aqueous solution ( c = 4.184 J/g⋅K ) if the reaction causes the temperature of the reservoir to rise from 22.0 ºC to 26.5 ºC ?
To calculate the heat transferred by the chemical reaction, we can use the equation:
q = mcΔT
where q is the heat transferred, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
Given:
m = 155 g
c = 4.184 J/g⋅K
ΔT = 26.5 ºC - 22.0 ºC = 4.5 ºC
Substituting these values into the equation, we get:
q = (155 g) x (4.184 J/g⋅K) x (4.5 ºC)
q = 29168.98 J or approximately 29.2 kJ
Therefore, the heat transferred by the chemical reaction to the calorimeter reservoir is 29.2 kJ.
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a mixture of 2.00 moles of h2, 3.0 moles of nh3 and 4.00 moles of co2 and 5.00 moles of n2 exerts a total pressure of 800 torr. what is the partial pressure of each gas?
The partial pressure of H in the mixture is 160 torr, 240 torr, 320 torr, and 400 torr, respectively.
The total pressure of the mixture is 800 torr. To calculate the partial pressure of each gas, you will need to use the ideal gas law equation, PV = nRT, where P is the pressure of the gas, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature.
Since the total pressure is constant, the equation can be rearranged as follows:
P1 = (n1/ntotal) x Ptotal = (n1/ntotal) x 800 torr.
Using this formula, we can calculate the partial pressure of each gas in the mixture:
Partial pressure of H2 = (2.00 moles / (2.00 + 3.00 + 4.00 + 5.00)) x 800 torr = 160 torrPartial pressure of NH3 = (3.00 moles / (2.00 + 3.00 + 4.00 + 5.00)) x 800 torr = 240 torrPartial pressure of CO2 = (4.00 moles / (2.00 + 3.00 + 4.00 + 5.00)) x 800 torr = 320 torrPartial pressure of N2 = (5.00 moles / (2.00 + 3.00 + 4.00 + 5.00)) x 800 torr = 400 torr
Therefore, the partial pressure of H in the mixture is 160 torr, 240 torr, 320 torr, and 400 torr, respectively.
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calculate a) the molality of ch3oh (methanol) and b) mole fraction of solvent in a solution that is 7.50% by mass ch3oh in ch3ch2oh (ethanol).
The molality of CH3OH is 0.03077 m and the mole fraction of CH3OH is 0.1326.
To calculate the molality of CH3OH (methanol) and the mole fraction of solvent in a solution that is 7.50% by mass CH3OH in CH3CH2OH (ethanol), we can use the following steps:
1. Calculate the moles of CH3OH present in the solution:
Mass of CH3OH = 7.50% by mass × 0.100 L solution = 0.00750 L CH3OH
Moles of CH3OH = 0.00750 L ÷ 24.3 g/mol = 0.0003077 mol CH3OH
2. Calculate the molality of CH3OH:
Molality of CH3OH = moles of CH3OH ÷ 0.100 L solution
= 0.0003077 mol ÷ 0.100 L = 0.03077 m
3. Calculate the moles of CH3CH2OH present in the solution:
Mass of CH3CH2OH = 100% - 7.50% = 92.50% by mass × 0.100 L solution = 0.09250 L CH3CH2OH
Moles of CH3CH2OH = 0.09250 L ÷ 46.1 g/mol = 0.002005 mol CH3CH2OH
4. Calculate the mole fraction of CH3OH:
Mole fraction of CH3OH = moles of CH3OH ÷ total moles
= 0.0003077 mol ÷ (0.0003077 mol + 0.002005 mol) = 0.1326
Therefore, the molality of CH3OH is 0.03077 m and the mole fraction of CH3OH is 0.1326.
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if 7.66 g of cuno3 is dissolved in water to make a 0.140 m solution, what is the volume of the solution in milliliters?
The volume of the solution in milliliters is 547.13 mL.
How to calculate the volume of the solution in milliliters?
The molarity of the solution is given by;
Molarity = Number of moles of solute / Volume of solution in liters
Using the above formula, we can calculate the volume of the solution as;
Volume of solution in liters = Number of moles of solute / Molarity
Number of moles of CuNO3 can be determined as follows:
Number of moles = Given mass of the substance / Molar mass of the substance
= 7.66 g / (Cu: 63.55 g/mol + N: 14.01 g/mol + 3O: 3 x 16 g/mol)
= 0.05 mol
Substituting the values of molarity and number of moles of CuNO3 in the formula of volume of solution, we get:
Volume of solution in liters = Number of moles of solute / Molarity
= 0.05 mol / 0.140 M = 0.357 L
Converting the volume in liters to milliliters;
Volume in milliliters = Volume in liters × 1000
= 0.357 L × 1000= 357 mL
Thus, the volume of the solution in milliliters is 357 mL.
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Measurements show that the energy of a mixture of gaseous reactants decreases by 370 kJ during a certain chemical reaction, which is earned out at a constant pressure. Furthermore, by carefully monitoring the volume change, it is determined that -97 kJ of work is done on the mixture during the reaction.
a. Calculate the change in the enthalpy of the gas mixture during the reaction. Round your answer to 3 significant digits.
b. Is the reaction exothermic or endothermic?
a.
Calculation of the change in the enthalpy of the gas mixture during the reaction
It is known that:ΔH = ΔE + PΔVWhere,
ΔH = Change in enthalpy of the gas mixture
ΔE = Change in internal energy of the gas mixture
P = PressureΔV = Change in volume of the gas mixture
Now, according to the problem statement,
E = -370 kJ/mol = constant (since the reaction is carried out at constant pressure) ΔV = -97 kJ
Substituting the values in the above formula: H = -370 kJ + constant (-97 kJ) H = -370 kJ; constant = 97 kJ ΔH = -370 kJ - 97 constant kJ
This is the required change in enthalpy of the gas mixture during the reaction, where the value of the constant is unknown.
b.
Is the reaction exothermic or endothermic?
The reaction is exothermic if the value of H is negative and endothermic if the value of H is positive.
As per the above calculation, the value of H is negative.
Therefore, the reaction is exothermic.
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