<span>2 NH</span>₃<span> + 3 O</span>₂<span> + 2 CH</span>₄<span> </span>⇒<span> 2 HCN + 6 H</span>₂<span>O
2mol : 2mol
34g : 54g
25,1g : x
x = (25,1g * 54g) / 34g </span>≈ 39,9g<span>
</span>
Answer:
9.4 liter
Explanation:
1) Data:
V₁ = 10.0 L
T₁ = 25°C = 25 + 273.15 K = 298.15 K
P₁ = 98.7 Kpa
T₂ = 20°C = 20 + 273.15 K = 293.15 K
P₂ = 102.7 KPa
V₂ = ?
2) Formula:
Used combined law of gases:
PV / T = constant
P₁V₁ / T₁ = P₂V₂ / T₂
3) Solution:
Solve the equation for V₂:
V₂ = P₁V₁ T₂ / (P₂ T₁)
Substitute and compuite:
V₂ = P₁V₁ T₂ / (P₂ T₁)
V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)
V₂ = 9.4 liter ← answer
You can learn more about gas law problems reading this other answer on
Explanation:
Explanation:
P1= 44 kpa
P2= 50 kpa
V1= 4.50 L
V2= ?
P1 V1= P2 V2
44 × 4.50 = 50 × V2
198= 50 × V2
V2 = 198/ 50
V2= 3.96 L "the new volume"
Answer:
m = 4450 g
Explanation:
Given data:
Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)
Initial temperature = 23.0°C
Final temperature = 57.8°C
Specific heat capacity of water = 1 cal/g.°C
Mass of water in gram = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 57.8°C - 23.0°C
ΔT = 34.8°C
4450 cal = m × 1 cal/g.°C × 34.8°C
m = 4450 cal / 1 cal/g
m = 4450 g
