Substances have a fixed arrangement of atoms. true or false (NO LINKS PLEASE)

What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin

Contact [7]

Answer: The total volume of the gaseous products is 1044.29 L

Explanation:

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = $\frac{1}{22.4}\times 116=5.18mol$ of butane

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

For carbon dioxide:

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = $\frac{8}{2}\times 5.18=20.72mol$ of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

For water vapor:

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = $\frac{10}{2}\times 5.18=25.9mol$ of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

3 0

3 years ago

Nicolaus Copernicus was a Polish astronomer who is best known for the theory that the sun is near the center of the universe and

Simora [160]

It is B nevermind I thought it was c but it wasn't.

5 0

3 years ago

PLEASE HELP WITH THIS QUESITION WILL GIVE BRAINLIST 2 BEST ANSWER.

Lera25 [3.4K]

Answer: show all

Explanation:

today is my b\day

4 0

2 years ago

The density of H2O2 is 1.407 g/mL, and the density of O2 is 1.428 g/L. How many liters of O2 can be made from 55 mL H2O2

balandron [24]

Explanation:

mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g

molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol

moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2

For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).

moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2

Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.

Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.

which is 53 mL.

8 0

2 years ago

a 160 milligram sample of a radioactive isotope decays to 10 kilograms in 12 years. what is the half life of this element

Shkiper50 [21]

Answer:

3 years

Explanation:

Given data:

Initial amount of sample = 160 Kg

Amount left after 12 years = 10 Kg

Half life = ?

Solution:

at time zero = 160 Kg

1st half life = 160/2 = 80 kg

2nd half life = 80/2 = 40 kg

3rd half life = 40 / 2 = 20 kg

4th half life = 20 / 2 = 10 kg

Half life:

HL = elapsed time / half life

12 years / 4 = 3 years

8 0

3 years ago