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oee [108]
3 years ago
7

A gas that was heated to 150 Celsius has a new volume of 1587.4 L. What was its volume when its temperature was 100 Celsius?

Chemistry
1 answer:
sdas [7]3 years ago
6 0

Answer:

\boxed {\boxed {\sf 1058.3 \ L}}

Explanation:

We are asked to find the new volume of a gas after a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The gas was heated to 150 degrees Celsius and had a volume of 1587.4 liters.

\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{T_2}

The temperature was 100 degrees Celsius, but the volume is unknown.

\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C}

We are solving for the volume at 100 degrees Celsius, so we must isolate the variable V₂. It is being divided by 100°C and the inverse of division is multiplication. Multiply both sides of the equation by 100°C.

100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C} * 100 \textdegree C

100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = V_2

The units of degrees Celsius cancel.

100  *\frac {1587.4 \ L }{150 } = V_2

100 *10.58266667  \ L = V_2

1058.266667 \ L = V_2

The original measurement of volume has 5 significant figures, so our answer must have the same. For the number we calculated, that is the tenth place. The 6 in the hundredth place to the right tells us to round to 2 up to a 3.

1058.3 \ L = V_2

The volume of the gas at 100 degrees Celsius is approximately <u>1058.3 liters.</u>

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Answer:

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Explanation:

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molarity = moles of solute ÷ liters of solution

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

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The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =
LuckyWell [14K]
To get the value of ΔG we need to get first the value of ΔG°:

when ΔG° = - R*T*㏑K

when R is constant in KJ = 0.00831 KJ

T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4

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then, we can now get the value of ΔG when:

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when ΔG° = -19 KJ

and R is constant in KJ = 0.00831 

and T is the temperature in Kelvin = 298 K

and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m  

so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

       = -40

     
8 0
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