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IceJOKER [234]
3 years ago
5

A container of what volume would be required to store 60.0 moles of a gas at a pressure of 20.0 atm and a temperature of 50.0 Ce

lsius? 12.3 L 79.6 L 8060 L 1250 L
Chemistry
1 answer:
belka [17]3 years ago
7 0

Answer:

The answer to your question is:  79.6 L

Explanation:

Data:

n = 60 moles

Pressure = P = 20.0 atm

Temperature = T = 50°C

Volume = V = ?

Convert temperature to Kelvin

      °K = 273 + 50

      °K = 323

Formula  (ideal gas law)

               PV = nRT        solve for V

              V = nRT / P

             V = (60)(0.08206)(323) / 20        Substitution

            V = 1590.32 /20                    

             V = 79.6 L

               

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2 years ago
Please help me like now please
777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

<u>Question 1</u>

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

<em>To balance the equation, count the number of atoms on both sides of the equation</em>

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

<em>Combining the pluses will give you the following</em>

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

<em>Both sides are the same, therefore the chemical equation is balanced (originally). </em>

From the equation, we can see that <u>1 mole of NaHCO3</u> produces <u>1 mole of CO2</u>.

So that means <u>0.5 mole of NaHCO3</u> would produce <u>0.5 mole of CO2</u>.

<u>Question 2</u>

C4H10 + O2 --> CO2 + H2O

<em>Again, count the number of atoms on both sides of the equation</em>

(4C, 10H, 2O) --> (1C, 2H, 3O)     <em>This time left does not equal right side</em>

<em>You now need to find </em><u><em>factors </em></u><em>that can make both sides equal. </em>

C4H10 + O2 --> <u>4</u>CO2 + H2O    <em>Now the C is balanced, let's recount </em>

<em>(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H)      H&O is still not balanced</em>

C4H10 + O2 --> 4CO2 + <u>5</u>H2O    <em>Now the H is balanced, let's recount</em>

<em>(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H)      O is still not balanced</em>

C4H10 + (<u>13/2</u>)O2 --> 4CO2 + 5H2O    <em>Now the O is balanced</em>

<em>(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)</em>

<em>But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).</em>

<u>2</u>C4H10 + <u>13</u>O2 --> <u>8</u>CO2 + <u>10</u>H2O    Now it's completely balanced!

<em>(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H)     Yayy! It's balanced.</em>

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

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A gas sample occupies 4.2L at a pressure of 101 kPa. What volume will it occupy if the pressure is increased to 235 kPa
taurus [48]

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<h3>Explanation</h3>

The volume of a gas, V, is inversely proportional to the pressure on it, P. That is:

V₁ · T₁ = V₂ · T₂.

Rearranging gives:

V₂ = V₁ · T₁ / T₂ = 4.2 × (101 / 235) = 1.8 L

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If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the
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Answer:

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Explanation:

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Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

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Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

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2 years ago
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