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Natasha2012 [34]
3 years ago
7

A sample of argon gas has a volume of 73 mL at a pressure of 1.20 atm and a temperature of 112 degrees Celsius. What is the fina

l volume of the gas, in mL, when the pressure and temperature of the gas sample are changed to .55 atm and 75 degrees C.
Chemistry
1 answer:
NeX [460]3 years ago
3 0

Answer:

V₂ → 106.6 mL

Explanation:

We apply the Ideal Gases Law to solve the problem. For the two situations:

P . V = n . R . T

Moles are still the same so → P. V / R. T = n

As R is a constant, the formula to solve this is: P . V / T

P₁ . V₁ / T₁ = P₂ .V₂ / T₂   Let's replace data:

(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C

((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂

58.66 mL.atm = 0.55 atm . V₂

58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL

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Answer:

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Explanation:

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What is combination reaction and also define it's properties?​
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Pablo drove 260 miles using 12 gallons of gas. at this rate, how many gallons of gas would he need to drive 286 miles?
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7 0
3 years ago
Calculate the mass of sodium azide required to decompose and produce 2.104 moles of nitrogen. Refer to the periodic table to get
Alexxx [7]

91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.

Explanation:

2NaN3======2Na+3N2

This  is the balanced equation for the decomposition and production of sodium azide required to produce nitrogen.

From the equation:

2 moles of NaNO3 will undergo decomposition to produce 3 moles of nitrogen.

In the question moles of nitrogen produced is given as 2.104 moles

so,

From the stoichiometry,

3N2/2NaN3=2.104/x

= 3/2=2.104/x

3x= 2*2.104

   = 1.4 moles

So, 1.4 moles of sodium azide will be required to decompose to produce 2.104 moles of nitrogen.

From the formula

no of moles=mass/atomic mass

        mass=no of moles*atomic mass

                   1.4*65

               = 91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.

4 0
3 years ago
I
Damm [24]

Answer: the answer is 18

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3 years ago
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