A sample of gas at 90 degrees Celsius occupies 15.5 L. This gas is heated to occupy a new volume of 20.3L. What is the new temperature of the gas?

202 °C

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 90 °C

Initial volume (V₁) = 15.5 L

Final volume (V₂) = 20.3 L

Final temperature (T₂) =?

Next, we shall convert 90 °C to Kelvin temperature. This is illustrated below:

T(K) = T(°C) + 273

Initial temperature (T₁) = 90 °C

Initial temperature (T₁) = 90 °C + 273

Initial temperature (T₁) = 363 K

Next, we shall determine the final temperature.

Initial temperature (T₁) = 363 K

Initial volume (V₁) = 15.5 L

Final volume (V₂) = 20.3 L

Final temperature (T₂) =?

V₁ / T₁ = V₂ / T₂

15.5 / 363 = 20.3 / T₂

Cross multiply

15.5 × T₂ = 363 × 20.3

15.5 × T₂ = 7368.9

Divide both side by 15.5

T₂ = 7368.9 / 15.5

T₂ ≈ 475 K

Finally, we shall convert 475 K to celsius temperature. This is illustrated below:

T(°C) = T(K) – 273

T₂ = 475 – 273

T₂ = 202 °C

Thus, the new temperature of the gas is 202 °C

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"