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2 years ago

How many molecules of Oxygen are consumed if 24 molecules of Carbon Dioxide are produced?

Chemistry

1 answer:

Nat2105 [25]2 years ago

6 0

Answer:

Explanation:

C + O2 → CO2

Mole of C = 24 g/(12 g/mole)

Mole of C = 2 mole

Mole of molecular O2 = 74 g/(32 g/mole)

Mole of molecular O2 = 2.3125 mole

Since mole of C < mole of O2, then C being the limiting reagent.

From the reaction, it shows that mole ratio between C and O2 = 1 : 1.

So, 2 moles of C will stoichiometrically react with 2 moles of O2 to generate 2 moles of CO2.

Avogadro's law states that :"equal volumes of all gases, at the same temperature and pressure, have the same number of molecules i.e. 6.02 x 10^23 molecules/mole.

Therefore, 2 moles of CO2 contain 2 moles x 6.02 x 10^23 molecules/mole = 1.204 x 10^24 molecules of CO2 is formed.

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A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for

laiz [17]

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 $= 10$ ml

Molar concentration of Ca(NO3)2 $= 0.001$

Volume of NaF $= 10$ ml

Molar concentration of NaF $= 0.0001$

Ksp for CaF2 $= 3.2 * 10^ {-11}$

CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2

Moles of calcium ion

$= 10 * 0.001\\= 0.01$

$[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}$

Moles of F- ion

$= 10 * 0.0001\\= 0.001$

$[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}$

$Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}$

Q is lesser than Ksp value of CaF2. Hence it will not precipitate

5 0

2 years ago

5. What is the mole fraction of NaOH in an aqueous<br> solution that contains 31.0 % NaOH by mass?

Liono4ka [1.6K]

Answer:

16.8%

Explanation:

31% NaOH molar mass 40 gm

69% H2O molar mass 18 gm

1000 gm would be

310 gm NaOH or 310/40 = 7.75 moles

690 gm of H2O or 690/18 = 38.333 moles

7.75 / (7.75 + 38.333) = .168 mole fraction

6 0

2 years ago

If the density and volume of an object is known, what can also be found?

ludmilkaskok [199]

The mass of the object can also be determined if the density and volume of an object are known.

8 0

3 years ago

Read 2 more answers

Type the correct answer in each box.

Arturiano [62]

Answer:

5SiO2 + 2CaC2 = 5Si + 2CaO + 4CO2

Explanation:

balancing equations is a lot of trial and error. My strategy to approaching this equation was to get the O's balanced. After trying several combonations I found that I needed 10 O's on each side of the equation for the other elements to match up. After I balanced the O's, I balanced my C's to 4 on each side. Then I balanced my Ca's to have 2 on each side. And last but not least I balanced my Si to have 5 on each side.

8 0

2 years ago

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0

2 years ago