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faust18 [17]
2 years ago
5

How many liters of oxygen (at STP) are required to form 10.5 grams of hydrogen

Chemistry
1 answer:
Bond [772]2 years ago
8 0

Answer:

You'd need 6.55 L of oxygen at STP to form that much water.

Explanation:

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An exhaled air bubble underwater at 290.
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The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 290 kPa

Final pressure (P_{2}) = 104 kPa

Initial volume (V_{1}) = 18.9 ml

To find:

Final volume (V_{2})

We know;

From the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

   P × V = constant

   P ∝ \frac{1}{V}

From the above equation;

              \frac{P_{1} }{P_{2} }  = \frac{V_{2} }{V_{1} }

P_{1} represents the initial pressure of the gas

P_{2} represents the final pressure of the gas

V_{1} represents the initial volume of the gas

V_{2} represents the final volume of the gas

Substituting the values of the above equation;

                    \frac{290}{104} = \frac{V_{2} }{18.9}

             V_{2} = 52.7 ml

<u><em>Therefore the final volume of the gas is 52.7 ml.</em></u>

6 0
2 years ago
Can someone help?? This is really hard.
Sergeu [11.5K]

Answer:

See Explanation

Explanation:

Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.

2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4

2^1 = 2^1

The rate of reaction is first order with respect to Hbn

Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.

1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4

3^1 = 3^1

The reaction is also first order with respect to CO

b) The overall order of reaction is 1 + 1=2

c) The rate equation is;

Rate = k [CO] [Hbn]

d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4  /[5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4/6.7 * 10^-7

k = 4.7 * 10^2 mmol-1 L s-1

e) The reaction occurs in one step because;

1) The rate law agrees with the experimental data.

2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.

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