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ziro4ka [17]
3 years ago
9

If 15 L of chlorine reacts at a constant temperature of 298 K and pressure of 9 atm, how many grams of hydrochloric acid (HCl) a

re produced?
Chemistry
1 answer:
stepan [7]3 years ago
7 0

Answer:400k

Explanation:

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Calculate the volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute. Enter your answer in the provided box. IL
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Answer:

The volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute is 0,25 L

Explanation:

Molarity (M) means: moles of solute which are contained in 1 L of solution.  

In this case we have 0,2 moles which are in 1 L, so, as we have 5x10*-2 moles we have to apply a rule of three to find out the volume.

0,2 moles ........... 1 L

5x10*-2 moles ........... x

x= (5x10*-2 moles . 1 L) / 0,2 moles = 0.25L

(we can also say 250 mL)

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Calculate the number of pounds of CO2 released into the atmosphere when a 22.0 gallon tank of gasoline is burned in an automobil
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Answer:

391.28771 pounds of carbon-dioxide released into the atmosphere.

Explanation:

Density of the gasoline ,d= 0.692 g/mL

Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL

Let mass of the gasolin be M

d=\frac{M}{V}

M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g

Assume that gasoline is primarily octane (given)

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

Mass of octane burnt in the tank =  M = 57,629.081 g

Moles of octane =\frac{57,629.081 g}{114.08 g/mol}=505.1637 mol

According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.

Then 505.1637 mol of octane will give:

\frac{16}{2}\times 505.1637 mol=4,041.3100 mol of carbon-dioxide

Mass of 4,041.3100 mol of carbon-dioxide:

4,041.3100 mol × 44.01 g/mol = 177,858.05 g

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