Answer:
Should be either mendele or fermium
maybe lawrencium not sure but I think it’s lawrencium! Sorry if wrong heh
Explanation: <u><em>really trying to help! If it’s wrong super sorry!!!</em></u>
You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles
Answer:
V₂ = 1070 mL or 1.07 L
Solution:
Data Given;
P₁ = 1170 mmHg
V₁ = 915 mL
T₁ = 24 °C + 273 K = 297 K
P₂ = 842 mmHg
V₂ = ?
T₂ = - 23 °C + 273 K = 250 K
According to Ideal gas equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = P₁ V₁ T₂ / P₂ T₁
Putting Values,
V₂ = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)
V₂ = 1070 mL or 1.07 L
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction? First you need to write a balanced chemical equation.
