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solong [7]
4 years ago
14

95.8 L of fluorine gas is being held at a temperature of 24.5oC. If the temperature were raised to 46.9 oC, what would the new v

olume be?
Chemistry
1 answer:
Xelga [282]4 years ago
6 0

Answer: 103 litres

Explanation:

Given that,

Original volume of fluorine gas (V1) = 95.8L

Original temperature of fluorine gas (T1) = 24.5°C

[Convert 24.5°C to Kelvin by adding 273

24.5°C + 273 = 297.5K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 46.9°C

[Convert 46.9°C to Kelvin by adding 273

46.9°C + 273 = 319.9K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

95.8L/297.5K = V2/319.9K

To get the value of V2, cross multiply

95.8L x 319.9K = 297.5K x V2

30646.42L•K = 297.5K•V2

Divide both sides by 297.5K

30646.42L•K/297.5K = 297.5K•V2/297.5K

103L = V2

Thus, the new volume of fluorine gas will be 103 litres

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5 0
3 years ago
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If a pork roast must absorb
Black_prince [1.1K]
Let us  assume propane was the fuel
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1 mole ofpropane produces 3 moles of CO2 
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7 0
3 years ago
What is the concentration of H+ ions at a pH = 11?
natta225 [31]

Answer:

\huge 1 × {10}^{-11} \: \: M

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ {H}^{+} ]

Since we are finding the H+ ions we find the antilog of the pH

So we have

11 =  -  log({H}^{+})  \\ {H}^{+} =  {10}^{ - 11}

We have the final answer as

1 × {10}^{-11} \: \: M

Hope this helps you

6 0
3 years ago
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Oxygen's molar mass is 16gm

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8 0
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