Answer:
80L
Explanation:
V1/T1 = V2/T2
V2 = V1 T2/T1
T1 = 300K
V1 = 60L
T2 = 400K
V2 = ?
V2 = V1 T2/T1
V2 = (60L)(400K) / (300K)
V2 = 80L
Answer:
363C
Explanation:
V = KT and T = V/K
3.5=k(35+273) = 308k
k=3.5/308 =0.011
new T=7.0/0.011=636K = 636-273 =363C
Answer:
The energy released will be -94.56 kJ or -94.6 kJ.
Explanation:
The molar mass of methane is 16g/mol
The given reaction is:
the enthalpy of reaction is given as ΔH = -890.0 kJ
This means that when one mole of methane undergoes combustion it gives this much of energy.
Now as given that the amount of methane combusted = 1.70g
The energy released will be:
Answer:
The answer to your question is 21.45 g of KBr
Explanation:
Chemical reaction
2K + Br₂ ⇒ 2KBr
14.4 ?
Process
1.- Calculate the molecular mass of bromine and potassium bromide
Bromine = 2 x 79.9 = 159.8g
Potassium bromide = 2(79.9 + 39.1) = 238 g
2.- Solve it using proportions
159.8 g of Bromine ------------ 238 g of potassium bromide
14.4 g of Bromine ------------ x
x = (14.4 x 238) / 159.8
x = 3427.2 / 159.8
x = 21.45g of KBr
