When hydrogen (H+) and chlorine (Cl-) are mixed, they are called hydrochloric acid.
How hydrochloric acid is formed?Hydrochloric acid is prepared by combining the gaseous hydrogen chloride with water. Due to the corrosive nature of the acid, ceramic, glass, or tantalum apparatus is used. Hydrochloric acid is also known as muriatic acid. It is also known as an aqueous solution of hydrogen chloride. It is a colorless solution with a pungent smell. It is classified as a strong acid because of more hydrogen atoms. It is a part of the gastric acid that is present in the digestive systems of animal species such as humans. Hydrochloric acid is an important laboratory and industrial chemical.
So we can conclude that hydrochloric acid is formed when hydrogen and chlorine are mixed.
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Ascorbic acid (vitamin C) is important in many metabolic reactions in the body, including the synthesis of collagen and prevention of scurvy. Given that themass percent composition of ascorbic acid is 40.9% C, 4.58% H, and 54.5% O, determine the empirical formula of ascorbic acid. Show all your work
Empirical formula:
Step 1
Information already provided
The mass percent composition:
40.9 % C
4.58 % H
54.5 % O
Information needed: from the periodic table
For C) 1 mol = 12.01 g
For H) 1 mol = 1.008 g
For O) 1 mol = 15.99 g
---------------------------
Step 2
A sample of 100 g is assumed, so:
40.9 % C => 40.9 g C
4.58 % H => 4.58 g H
54.5 % O => 54.5 g O
--------------------------
Step 3
Convert mass into moles:
40.9 g C x (1 mol/12.01 g) = 3.40 moles C
4.58 g H x (1 mol/1.008 g) = 4.54 moles H
54.5 g O x (1 mol/15.99 g) = 3.40 moles O
------------------------
Step 4
All moles calculated in step 3 need to be divided by the smallest one.
3.40 moles C/3.40 moles = 1
4.54 moles H/3.40 moles = 1.33
3.40 moles O/3.40 moles = 1
-----------------------
Step 5
Integer numbers are needed, so let's multiply by 3 all of them in step 4
Therefore,
For C) 3
For H) 3.99 = 4 approx.
For O) 3
All these numbers calculated will be the subindexes in ascorbic acid
Answer:
Empirical formula: C3H4O3
3. If you need to produce 85 g of CO2, how many grams of: (these are 3 problems starting with thea. C3H8, do you need?same amount:b. O2, do you need?c. H2O will also be made?
1) First let's write the equation. It is a combustion reaction, so:
C₃H₈ + O₂ ---> CO₂ + H₂O
and balance the equation (same number of atoms of each element on both sides of the equation):
C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O
Reactant side:
C - 3
H - 8
O - 10
Product side:
C - 3
O - 10
H - 8
2) Now let's transform 85 grams of CO₂ into mole. For this, we use the following equation:
mole = mass/molar mass
Molar mass of CO₂ is: (1×12) + (2×16) = 44 g/mol
mole = 85/44
mole = 1.9 mol of CO₂
3) Now we use the proportion of the balanced equation:
1 mol of C₃H₈ ---- 3 mol of CO₂
x mol of C₃H₈ ----- 1.9 mol of CO₂
x = 0.6 mol of C₃H₈
4) Now we transform mole of C₃H₈ into grams using its molar mass.
molar mass of C₃H₈ is: (3×12) + (8×1) = 44 g/mol
mass = mole × molar mass
mass = 0.6 × 44
mass of C₃H₈ = 28 g
Answer: a) mass of C₃H₈ = 28 g
For alternative b we follow the same process starting from step 3:
3)Now we use the proportion of the balanced equation:
5 mol of O₂ ---- 3 mol of CO₂
x mol of O₂ ----- 1.9 mol of CO₂
x = 3.16 mol of O₂
4) Now we transform mole of O₂ into grams using its molar mass.
molar mass of O₂ is: (2×16) = 32 g/mol
mass = mole × molar mass
mass = 3.16 × 32
mass of O₂ = 101 g
Answer: b) mass of O₂ = 101 g
For alternative c we follow the same process starting from step 3:
3) Now we use the proportion of the balanced equation:
4 mol of H₂O ---- 3 mol of CO₂
x mol of H₂O ----- 1.9 mol of CO₂
x = 0.84 mol of H₂O
4) Now we transform mole of H₂O into grams using its molar mass.
molar mass of H₂O is: (2×1) + (1×16) = 18 g/mol
mass = mole × molar mass
mass = 0.84 × 18
mass of H₂O = 15 g
Answer: c) mass of H₂O = 15 g
The value of AH rxn for the following reaction is -72 kJ. How many kJ of heat is released when 0.989 g of HBr (80.91 g/mol) is formed? H2 (g) + Br2 (g) -› 2 HBr (gram). A. -144B. -72 C. -0.44 D. -36
Answer:
[tex]C\text{ : -0.44 KJ}[/tex]Explanation:
Here, we want to get the amount of heat released in KJ
From the change in enthalpy given and the equation of reaction, we know that 2 moles of HBr would lead to that amount of heat
Now, let us get the actual amount of heat released
We need to get the actual number of moles of HBr produced
Mathematically, we can calculate that by dividing the mass of HBr by its molar mass
We have that as:
[tex]\frac{0.989}{80.91}\text{ = 0.0122 mol}[/tex]From the reaction information:
-72 KJ was released by 2 moles
x KJ would be released by 0.0122 mol
To get the value of x, we have it that:
[tex]\begin{gathered} x\text{ }\times2\text{ = 0.0122 }\times\text{ \lparen-72\rparen} \\ \\ x\text{ = -36 }\times\text{ \lparen0.0122\rparen = -0.44 KJ} \end{gathered}[/tex]A gas occupying 0.6 L at 1.70 atm expands to 0.9 L. What is the new pressure assuming temperature remains constant?
Answer:
1.13 atmExplanation:
The new pressure can be found by using the formula
P1V1 = P2V2
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we're finding the new pressure P2 we make P2 the subject
We have
[tex]p_2 = \frac{p1v1}{v2} \\ [/tex]
P1 = 1.7 atm
V1 = 0.6 L
V2 = 0.9 L
We have
[tex]p_2 = \frac{1.7 \times 0.6}{0.9} = 1.13333...\\ [/tex]
We have the final answer as
1.13 atmHope this helps you
Can someone help me to answer this?As you reflect on how to interpret a balanced chemical reaction.Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 2002, Philippines has banned imports of ammonium nitrate that used in bombs that killed 12 people in Mindanao area.The explosion resulted from this reaction:2NH4NO3(s)→2N2(g)+4H2O(g)+O2(g)Construct a table showing how to interpret the information in the equation in terms of:1. individual molecules and ions.2. moles of reactants and products.3. grams of reactants and products given 2 mol of ammonium nitrate.4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.
(1)
The molecule NH4NO3 is ammonium nitrate, which makes a redox reaction.
It produces N2, O2, and H2O, which are nitrogen, oxygen, and water.
The ions are 2NH4+ and 2NO3-.
2 molecules of N2, 4 molecules of H2O, and 1 molecule of O2.
(2)
There are 2 moles of NH4NO3, 2 moles of N2, 4 moles of H2O, and 1 mole of O2.
(3)
The molar mass of NH4NO3 is 80.043 grams per mole, but there are 2 moles of it, so there are 160.09 grams of NH4NO3.
There are 56.03 grams of N2 because there are 2 moles of it. (1 mole N2 = 28.0134 g/mol).
There are 72.06 grams of H2O because there are 4 moles of it (1 mole H2O = 18.02 g/mol).
There are 31.998 grams of O2 because there's just 1 mole of it.
(4)
The formula units of NH4NO3 is 1.204x10^24, which is equivalent to 2 moles.
Nitrogen has the same formula units because there are 2 moles of it, so it's 1.204x10^24.
Water has 2.409x10^24 because there are moles of it.
Oxygen has 6.022x10^23 because there's just 1 mole, inis Avogadro's Number.
A student finds a piece of metal and finds its mass to be 750 g. Through water displacement thestudent determines the volume to be 65.8 cm". Which metal does the student have?
Metals normally have different densities, so we can try to determine which metal is this by its density.
Assuming it is pure, the density of the metal is its mass divided by its volume:
[tex]\begin{gathered} \rho=\frac{m}{V} \\ \rho=\frac{750g}{65.8cm^3} \\ \rho\approx11.4g/cm^3 \end{gathered}[/tex]Now, we need to look for some table with densities of various metals and see which one has the density we found, 11.4 g/cm³.
Since we don't have one, we can look for one. In it, we can see that the only metal with this density is lead.
So, the metal in the question should be lead.
54000 mL isO 54 m3O 5400 cm3O 0.054m3O 540 m3
mL and cm³ have a 1 to 1 conversion, so we have:
[tex]54000mL=54000cm^{3}[/tex]But we don't have this option, so we will need to find another.
The other are in m³, so we can use the conversion from cm to m to get this:
[tex]\begin{gathered} 1m=100cm \\ (1m)^{3}=(100cm)^{3} \\ 1m^{3}=1000000cm^{3} \\ 1cm^{3}=\frac{1}{1000000}m^{3} \end{gathered}[/tex]So, we can apply this to what we have:
[tex]54000cm^3=54000\cdot\frac{1}{1000000}m^3=0.054m^{3}[/tex]We have an option with 0.054m³, so the correc alternative is 0.054 m³.
The following lists consists of ionic compounds EXCEPT
barium hydroxide, zinc carbonate, ammonium sulfate
calcium chloride, carbon disulfide, magnesium nitrate
sodium sulfate, copper(II) oxide, potassium nitride
aluminium sulfide, sodium sulfite, calcium fluoride
The following lists consists of ionic compounds except carbon disulfide (CS₂).
Barium hydroxide , Ba(OH)₂ is an ionic compound.
zinc carbonate, ZnCO₃ is an ionic compound.
ammonium sulfate , (NH₄)₂SO₄ is an ionic compound.
calcium chloride, CaCl₂ is an ionic compound.
carbon disulfide, CS₂ is not an ionic compound. In carbon disulfide both the elements are non metallic elements. The bond formed between atoms are by sharing of electron known as covalent bond due to very little difference in electronegativity.
magnesium nitrate, Mg(NO₃)₂ is an ionic compound.
sodium sulfate, Na₂SO₄ is an ionic compound.
copper(II) oxide, CuO is an ionic compound.
potassium nitride KNO₃ is an ionic compound.
aluminium sulfide, Al₂S₃ is an ionic compound.
sodium sulfite, Na₂S is an ionic compound.
calcium fluoride, CaF₂ is an ionic compound.
Thus, The following lists consists of ionic compounds except carbon disulfide (CS₂).
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17 was found from determination in a mass spectrometer that an element X has three Isotopes whose mass are & 19.19,20.99 and 21.99 respectively The abundance of these I sotopes are 90.92% 0.25% $8.83% respectively Calculate the relative atomic mass.
The relative atomic mass of the given element is 40.372 amu.
What is relative atomic mass?The relative atomic mass of an element is considered as the sum of the isotopes masses each multiplied by the percentage which is found in nature.
The formula which is used to calculate the relative atomic mass is
Relative atomic mass = sum of all atomic masses of isotopes × fractional abundance
Given,
Mass of isotopes 1 = 19.19 amu
Mass of isotopes 2 = 20.99 amu
Mass of isotopes 3 = 21.99 amu
Fractional abundance of isotope 1 = 0.9092
Fractional abundance of isotope 2 = 0.0025
Fractional abundance of isotope 3 = 0.0883
By substituting all the values, we get
[( 19.19 × 0.9092) + (20.99 × 0.0025) + (21.99 × 0.883)]
= 17.447 + 0.052 + 22.873
= 40.372 amu.
Thus, we concluded that the relative atomic mass of the given element is 40.372 amu.
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What is the density of hydrogen sulfide (H2S) at 0.2 atm and 311 K?Answer in units of g/L
Answer
Density = 0.267 g/L
Explanation
Given:
Pressure of H2S = 0.2 atm
Temperature = 311 K
We know:
The molar mass of H2S = 34,1 g/mol
R constant = 0.08206 L.atm/K.mol
Solution:
From the ideal gas law:
PV = nRT
We know that:
density = m/V
n = m/M
Therefore we can use the following equation to solve for density of H2S
[tex]\begin{gathered} density\text{ = }\frac{PM}{RT} \\ density\text{ = }\frac{(0.2\text{ atm x 34,1 g/mol\rparen}}{(0.08206\text{ }L.atm/K.mol\text{ x 311 K\rparen}} \\ \\ density\text{ = 0.267 g/L} \end{gathered}[/tex]What is the largest possible electronegativity difference for a bond to be covalent?A.0.5B.1.7C.0.0D.1.0
Answer
B. 1.7
Explanation
As a rule, an electronegativity difference of 2 or more on the Pauling scale between atoms leads to the formation of an ionic bond. A difference of less than 2 between atoms leads to covalent bond formation.
Therefore, the largest possible electronegativity difference for a bond to be covalent is 1.7
Find the element that is oxidized and the one that is reduced Si + 2 F2 --> SiF4
Answer
The element that is oxidized is Si and the one that is reduced is F₂.
Explanation
Si + 2F₂ → SiF₄
The given reaction is an oxidation-reduction (redox) reaction:
Oxidation: Si → Si⁴⁺ + 4e⁻
Reduction: 2F₂ + 4e⁻ → 4F⁻
Si is a reducing agent, and F₂ is an oxidizing agent.
An oxidizing agent gains electrons and is reduced in a chemical reaction.
A reducing agent loses electrons and is oxidized in a chemical reaction.
Therefore, the element that is oxidized is Si and the one that is reduced is F₂.
At 302 K , to what pressure can the carbon dioxide in the cartridge inflate a 3.05 L mountain bike tire? (Note that the gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi .)
Ideal gas law is valid only for ideal gas not for vanderwaal gas. Therefore the carbon dioxide in the cartridge inflate a 3.05 L mountain bike tire to 104.3psi pressure.
What is ideal gas equation?Ideal gas equation is the mathematical expression that relates pressure volume and temperature.
Mathematically,
PV=nRT
where,
P = pressure
V= volume=3.05 L
n =number of moles=1mole(assumed as it is not given in question)
T =temperature = 302 K
R = Gas constant = 0.0821 L.atm/K.mol
P × 3.05 L =1 mole× 0.0821 L.atm/K.mol × 302 K
P =8.12atm
1 atm = 14.7 psi
Hence Pressure in psi = 8.12atm×14.7 psi = 119.49 psi
Pressure by the gas= Total pressure - Atmospheric pressure = 119.49 - 14.7 psi = 104.3psi
Therefore the carbon dioxide in the cartridge inflate a 3.79 L mountain bike tire to 104.3psi pressure.
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17. Which of the following represents a formula for a chemical compound?A. CB. KOHC. O
KOH. Option B is correct
Explanations:A chemical compound are made up of more than one element combined together. According to the question, we need to determine the formula that represents a compound.
The compound there is KOH since it contains three elements (Potassium, Oxygen and Hydrogen)
A reaction experimentally yields 15.68 g of a product. What is the percent yield if the theoretical yield is 18.81 g?
Answer
The percent yield = 83.36%
Explanation
Given:
Experimental yield = actual yield = 15.68 g
Theoretical yield = 18.81 g
What to find:
The percent yield for the reaction.
Step-by-step solution:
The percent yield for the reaction can be calculated using the formula below:
[tex]\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{15.68\text{ }g}{18.81\text{ }g}\times100\% \\ \\ Percent\text{ }yield=83.36\% \end{gathered}[/tex]Hence, the percent yield for the reaction is 83.36%
I’m not sure and I’m kind of confused can anyone help?
We will reconstruct the model in the following manner :
From the above diagram we can see that :
• number of Carbon atom = 3
• number of hydrogen atom = 8
• rewrite this in an alphabetical order, you get :
[tex]\begin{gathered} C_3H_8\text{ } \\ \Rightarrow Propane\text{ } \end{gathered}[/tex]the molecule has a chemical formula = C3H8The quantatum mechanical model of an atom uses atomic orbitals to describe what
How many molecules of ethane gas, C2H6 are in 15 grams of the compound?
3.01×10²³molecules.
Explanations:
The formula for the number of molecules of a compound given the number of moles is expressed as:
[tex]nu\text{mber of molecules=moles}\times6.02\times10^{23}[/tex]Get the moles of ethane gas using the formula:
[tex]\begin{gathered} \text{moles of ethane=}\frac{Mass\text{ of ethane}}{Molar\text{ mass of ethane}} \\ \text{Moles of ethane=}\frac{15}{2(12)+1(6)} \\ \text{Moles of ethane}=\frac{15}{30}\text{moles} \\ \text{Moles of ethane}=0.5\text{moles} \end{gathered}[/tex]Determine the required number of molecules of ethane
[tex]\begin{gathered} nu\text{mber of mol}ecules=0.5\times6.02\times10^{23} \\ nu\text{mber of mol}ecules=3.01\times10^{23} \end{gathered}[/tex]Hence the molecule of ethane gas that is in 15 grams of the compound is 3.01×10²³molecules.
Classify CH3CH2NH2 as astrong base or a weak base.Strong BaseWeak Base
Answer:
CH3CH2NH2 is a weak base.
Explanation:
CH3CH2NH2 is a weak base since its Kb is small, and thus it partially dissociates.
Methane(CH4) gas and oxygen (O2) gas react to form carbon dioxide (CO2) gas and water vapor(H2O). Suppose you have 5.0 mol of CH4 and 1.0 mol of O2 in a reactor.
What would be the limiting reactant? Enter its chemical formula below
The limiting reactant, given that 5.0 moles of CH₄ and 1.0 mole of O₂ are in the reactor is O₂
How do I determine the limiting reactantWe'll begin by obtainig the balanced equation for the reaction. This is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
The limiting reactant for the reaciont can be obtained as illustrated below:
From the balanced equation above,
1 mole of CH₄ reacted with 2 moles of O₂
Therefore,
5 moles of CH₄ will react with = 5 × 2 = 10 moles of O₂
From the above illustration, we can see that a higher amount of O₂ is needed to react completely with 5 moles of CH₄.
Thus, we can conclude that O₂ is the limiting reactant.
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What functional group is found in amino acids?A) aminesB) alkanesB) cyclic hydrocarbon ringsC) alcohols
Answer:
[tex]A)\text{ Amines}[/tex]Explanation:
Here, we want to get the functional group is found in amino acids
In the amino acids, we can get amines, carboxylic acid groups, and the carbon chain
Looking at the options, we can see that the amine group is the right option here as the other two are not available
Determine the percent composition of hydrogen for the following: NaHCO3
By definition, the percent composition of an atom in a compound is its mass percentage in the formula.
That is, if we have 1 mol of NaHCO₃, we have also 1 mol of H (because there is only on H for each molecule).
So, we calculate the mass of this 1 mol of NaHCO₃ and the mass of 1 mol of H and calculate the percentage.
In equations, we want the following:
[tex]C_H=\frac{m_H}{m_{NaHCO_{3}}}[/tex]Since these are ratios, we doesn't matter if we talk about 1, 2 or any number of moles, but 1 mol is easier because the molecular and atomi masses are for 1 mol.
The molecular mass of NaHCO₃ is:
[tex]\begin{gathered} M_{NaHCO_3}=M_{Na}+M_H+M_C+3\cdot M_O \\ M_{NaHCO_3}\approx(22.990+1.008+12.011+3\cdot15.999)g/mol \\ M_{NaHCO_3}\approx84.006g/mol \end{gathered}[/tex]Which means that we have approximately 84.006 grams of NaHCO₃ in 1 mol of it.
The atomic mass of H is:
[tex]M_H\approx1.008g/mol[/tex]Which means that we have approximately 1.008 grams of H in 1 mol of it.
Now, we can take the percentage of mass of H:
[tex]C_H\approx\frac{1.008g_{}}{84.006g}\cdot100\%\approx1.20\%[/tex]So, the percentage composition of H in NaHCO₃ is approximately 1.20%.
If you wanted to dilute the 3M NaOH solution to 500mL of 1M NaOH solution, how much L of the 3M NaOH solution would you need?
0.167L
Explanation:In order to know how much L of the 3M NaOH solution would you need, we will simply set up an equal proportion solution expressed as;
[tex]C_1V_1=C_2V_2[/tex]C1 and C2 are the concentration of the solutions
V1 and V2 are the volumes of the solutions
Given the following parameters;
C1 = 1M
V1 = 500mL
C2 = 3M
V2 = ?
Substitute the given parameters into the formula above to get the required litre of solution needed.
[tex]\begin{gathered} 1\times500=3\times V_2_{} \\ 500=3V_2 \\ \text{Swap} \\ 3V_2=500 \end{gathered}[/tex]Divide both sides by 3:
[tex]\begin{gathered} \frac{3V_2}{3}=\frac{500}{3}_{} \\ V_2=166.67mL \end{gathered}[/tex]Converting to litres
Since 1mL = 0.001L
166.67mL = x
Cross multiply
[tex]\begin{gathered} 1\times x=166.67\times0.001 \\ x=0.167L \end{gathered}[/tex]Hence the amount of L of the 3M NaOH solution would you need is 0.167L
A solution with a total volume of 1000.0 mL contains 37.1 g Mg(NO3)2. If you remove 20.0 mL of this solution and then dilute this 20.0 mL sample with water until the new volume equals 500.0 mL, what is the concentration of Mg+2 ion in the 500.0 mL of solution? What is the concentration of nitrate ion?
1. The concentation of the magnesium ion, Mg²⁺ in the solution is 0.01 M
2. The concentation of the nitrate ion, NO₃⁻ in the solution is 0.02 M
We'll begin by obtaining the concentration of the stock solution. This can be obtained as follow:
Mass of Mg(NO₃)₂ = 37.1 gMolar mass of Mg(NO₃)₂ = 148 g/moleMole of Mg(NO₃)₂ = 37.1 / 148 = 0.25 moleVolume = 1000 mL = 1000 / 1000 = 1 LConcentration =?Concentration = mole / volume
Concentration = 0.25 / 1
Concentration = 0.25 M
Next, we shall determine the concentration of the diluted solution
Volume of stock solution (V₁) = 20 mLConcentration of stock solution (C₁) = 0.25 MVolume of diluted solution (V₂) = 500 mL Concentration of diluted solution (C₂) =?C₁V₁ = C₂V₂
0.25 × 20 = M₂ × 500
5 = M₂ × 500
Divide both side by 500
C₂ = 5 / 500
C₂ = 0.01 M
1. How to determine the concentration of magnesium ion, Mg²⁺
Mg(NO₃)₂(aq) <=> Mg²⁺(aq) + 2NO₃⁻(aq)
From the balanced equation above,
1 mole of Mg(NO₃)₂ contains 1 mole of Mg²⁺
Therefore,
0.01 M Mg(NO₃)₂ will also contains 0.01 M Mg²⁺
Thus, the concentration of Mg²⁺ is 0.01 M
2. How to determine the concentration of nitrate ion, NO₃⁻
Mg(NO₃)₂(aq) <=> Mg²⁺(aq) + 2NO₃⁻(aq)
From the balanced equation above,
1 mole of Mg(NO₃)₂ contains 2 mole of NO₃⁻
Therefore,
0.01 M Mg(NO₃)₂ will contain = 0.01 × 2 = 0.02 M NO₃⁻
Thus, the concentration of NO₃⁻ is 0.02 M
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8. Based on the Law of Conservation of Matter: At the start of the reaction 20g of
one material and some amount of another material were reacted and produced
30g of solid and 70g of a gas. What is the other amount of reactant used?
a. 20
b. 60g
c. 80g
d. 100g
Answer:
80g
Explanation:
The law of conservation of matter states that matter cannot be created nor destroyed. Basically, whatever mass you have at the beginning of a reaction, should be the same as at the end of the reaction on the product side.
Since here it lets you know that a total of 100g (30g + 70g) were produced on the product side, that means that we started off with 100g in our reactant side.
We have given that one material is 20g on our reactant side but we need the mass of the other. To find the mass of the other material, simply subtract 20g from the total mass created on the product side.
100g - 20g = 80g
The 80g would be the missing amount from the reactant side that isn't stated.
Substance Density (grams/cm3)Chloroform - 1.5Ebony wood - 1.2Mahogany wood - 0.85Oil - 0.9Water - 1.025.Since volume = mass/density, a 1,700 gram beam of mahogany wood has a volume of...Volume = Mass / DensitySelect one:a. 500 cm3b. 1,445 cm3c. 1,785 cm3d. 2,000 cm3
As the question gave us the formula in which we have to use to calculate the volume of this type of wood:
V = m/d
We have:
m = 1700 grams
d = 0.85
Now we add these values into the formula:
V = 1700/0.85
V = 2000 cm3, letter D
Name three different forms of mixture
Answer:
Mixtures can be classified on the basis of particle size into three different types: solutions, suspensions and colloids. The components of a mixture retain their own physical properties.
Sorry for the bad English, love from Vanuatu!
Explanation:
How many grams of H2 are required to completely convert 80g of Fe2O3?
Answer
3.0 grams H₂ is required.Explanation
Given:
Mass of Fe2O3 = 80 g
Equation:
What to find:
The grams of H2 required to completely convert 80g of Fe2O3.
Step-by-step solution:
From the equation of reaction;
3 moles of H2 completely react with 1 mole of Fe2O3
Note: Molar mass of H2 is 2.016 grams per mole and Molar mass of Fe2O3 is 159.69 g/mol
This implies; (3 x 2.016 g) = 6.048 grams H2 completely react with 159.69 grams Fe2O3.
Therefore, x grams H2 will completely convert 80 grams Fe2O3.
Cross multiply and divide both sides by 159.69 grams Fe2O3.
x grams H2 is now equal to
[tex]x=\frac{80\text{ }g\text{ }Fe_2O_3}{159.69\text{ }g\text{ }Fe_2O_3}\times6.048\text{ }g\text{ }H_2=3.0298\approx3.0\text{ }grams\text{ }H_2[/tex]Therefore the grams of H2 required to completely convert 80g of Fe2O3 is 3.0 grams
What impact did cargo ship refrigeration systems have on the banana industry? Refrigeration slows the rate at which food is being spoiled, refrigeration does not affect speed at which the ship moves, refrigeration does not cause ripening, refrigeration does not control the sugar content of bananas?
Answer
Refrigeration slows the rate at which food is being spoiled
Explanation
One of the importance of storing foods at cold temperatures (refrigeration) is to slow the growth of microorganisms, thereby limiting food poisoning while preserving food's nutritional qualities and good taste.
Therefore, the impact the cargo ship refrigeration systems have on the banana industry is:
Refrigeration slows the rate at which food is being spoiled
Describe global influences on local
weather.