The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L
From the question given above, the following data were obtained:
Initial pressure (P₁) = 8.5 atm
Initial volume (V₁) = 24 L
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Final pressure (P₂) = 13.5 atm
Final temperature (T₂) = 15 °C = 15 + 273 = 288 K
<h3>Final volume (V₂) =? </h3>
- The final volume of the gas can be obtained by using the combined gas equation as illustrated below:
Cross multiply
298 × 13.5 × V₂ = 204 × 288
4023 × V₂ = 58752
Divide both side by 4023
<h3>V₂ = 15 L </h3>
Therefore, the final volume of the gas is 15 L
Learn more: brainly.com/question/25547148
The number of grams of radon 222 did it have 15.2 ago was 49.6 grams( answer C)
<u>calculation</u>
- calculate the number of half life it has covered from 15.2 days to 3.8 days
that is divide 15.2/ 3.8 = 4 half life
- half life is time taken for a radio activity of a specified isotope to fall to half its original mass
therefore 3.8 days ago it was 3.1 x2 = 6.2 grams
7.6 days ago it was 6.2 x2 = 12.4 grams
11.4 days ago it was 12.4 x2= 24.8 grams
15.2 days ago it was 24.8 x2=49.6 grams
Mass of KCl= 1.08 g
<h3>Further explanation</h3>
Given
1 g of K₂CO₃
Required
Mass of KCl
Solution
Reaction
K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂
mol of K₂CO₃(MW=138 g/mol) :
= 1 g : 138 g/mol
= 0.00725
From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :
= 2/1 x mol K₂CO₃
= 2/1 x 0.00725
= 0.0145
Mass of KCl(MW=74.5 g/mol) :
= mol x MW
= 0.0145 x 74.5
= 1.08 g
Answer:
C = 0.2349 J/ (g °C)
Explanation:
Mass, m = 894.0g
Initial Temperature = −5.8°C
Final Temperature = 17.5°C
Temperature change = 17.5°C - (−5.8°C) = 23.3
Heat, H = 4.90kJ = 4900 J
Specific heat capacit, C = ?
The relationship between these quantities is given by the equation;
H = mCΔT
C = H / mΔT
C = 4900 / (894)(23.3)
C = 0.2349 J/ (g °C)
