- C_5H_8+13/2O_2—»5CO_2+4H_2O
Balanced one
- 2C_5H_8+13O_2—»10CO_2+8H_2O
Moles of Pentyne
- Given mass/Molarmass
- 34/68
- 0.5mol
Moles of H_2O
1mol releases 241.8KJ
2mol releases 241.8(2)=483.6KJ
Answer:
Q.89
Alkane - CnH(2n+2)
given that 8 H = > 8= 2n+2
therefore n= 3
C3H8 = 12×3 + 8×1= 36 +8 = 44
The Kelvin Scale is used, so the answer is temperature :)
Answer:
i know the answer the answer is valence electrons.
Answer: -
100 mm Hg
Explanation: -
P 1 =400 mm Hg
T 1 = 63.5 C + 273 = 336.5 K
T 2 = 34.9 C + 273 = 307.9 K
ΔHvap = 39.3 KJ/mol = 39.3 x 10³ J mol⁻¹
R = 8.314 J ⁻¹K mol⁻¹
Now using the Clausius Clapeyron equation
ln (P1 / P2) = ΔHvap / R x (1 / T2 - 1 / T1)
Plugging in the values
ln (400 mm/ P₂) = (39.3 x 10³ J mol⁻¹ / 8.314 J ⁻¹K mol⁻¹) x (
- 
= 1.38
P₂ = 100 mm Hg