In a closed system at 40°C, a liquid has a vapor

0.0145 moles of helium gas are introduced into a balloon so that the volume of the balloon is 2.54 liters. An additional amount

olga_2 [115]

Answer:

4.43L is final volume of the ballon

Explanation:

Avogadro's law of ideal gases states that equal volumes of gases, at the same temperature and pressure, have the same number of molecules.

The formula is:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

Where V and n are volume and moles of the gas in initial and final conditions.

If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:

$\frac{2.54L}{0.0145mol} =\frac{V_2}{0.0253mol}$

V₂ = 4.43L is final volume of the ballon

6 0

3 years ago

(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum

ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

4 0

1 year ago

Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en

Bezzdna [24]

Answer:

10.19 g CO₂

4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following conversion factors:

1 gallon = 3.785 L

1 L = 1000 mL

Now we convert 1.24 gallons to mL:

1.24 gallon * $\frac{3.785L}{1gallon} *\frac{1000mL}{1L} =$ 4693.4 mL

We calculate the mass of Octane:

4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the stoichiometric ratios and molecular weights to calculate the mass of CO₂ and H₂O:

CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{8molCO_{2}}{1molOctane}$ * 44 g/mol = 10.19 g CO₂

H₂O ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{9molH_{2}O}{1molOctane}$ * 18 g/mol = 4.69 g H₂O

7 0

3 years ago

50 POINTS*** Which of the following is not a correct chemical equation for a double displacement reaction?

hammer [34]

I mostly believe in between D and B beacuse K3po4 and caco3 is not an element equation

4 0

3 years ago

Read 2 more answers

Is this statement true or false? Bases feel slippery. A. True B. False

Ksenya-84 [330]

Actually, it is true

7 0

3 years ago

Read 2 more answers