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krek1111 [17]
3 years ago
12

5. What is the final "Celsius" temperature if 2.40 L of gas at 30.5 C is cooled until the volume reaches 1.00 L at constant pres

sure?
Chemistry
1 answer:
slega [8]3 years ago
7 0

Answer:

Final temperature of the gas = -146.63 °C

Explanation:

At constant pressure, volume and temperature of the gases are related as:

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

Where,

V1 = Initial volume = 1.00 L

V2 = Final volume = 2.40 L

T1 = Initial temperature = 30.5 °C = 30.5 + 273.15 = 303.65 K

Now, substitute the values in the above equation,

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

\frac{2.40\;L}{303.65}=\frac{1.00\;L}{T_2}

T_2=\frac{1.00\times 303.65}{2.40}

T2 = 126.52 K

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

T( °C) = T(K) - 273.15

          = 126.52 - 273.15 = -146.63 °C

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Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from
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Answer:

1.82 L

Explanation:

We are given the following information;

  • Initial volume as 2.0 L
  • Initial temperature as 60.0°C
  • New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

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

Where, V_1 and V_2 are initial and new volume respectively, while T_1 and T_2 are initial and new temperatures respectively;

T_1= 333 K

T_2=303K

V_1 =2.0L

Rearranging the formula;

V_2=\frac{V_1T_2}{T_1}

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Therefore, the new volume that would be occupied by the gas is 1.82 L

7 0
3 years ago
(-5)+(+7)-(-4) + (+12)<br>A 8<br>B 10<br>C 18<br>d 20<br>​
Vladimir79 [104]

Answer:

C-18

Explanation:

Step one follow order of operations

Add and subtract from left to right(-5)+(7)=2-(-4)+(12)

STEP 2

Apply negative Rule -(-4)=+4=2+4+(12)

then add 2+4+12=18

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