Answer:
Explanation:
2 moles hydrogen reacts with one mole of oxygen to give 2 moles of water.
a ) rate of consumption of hydrogen ( moles per second) is twice the rate of consumption of oxygen .
b ) rate of formation of water ( moles per second ) is twice the rate of consumption of oxygen
c ) rate of formation of water ( moles per second ) is equal to the rate of consumption of hydrogen.
The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams
calculation
lambda㏑2/18= 0.0385
m(t)= 165 x e( 0.0385 x90) =5.16g
Answer:
B
101L
Explanation:
We use the ideal gas relation
PV = nRT
P = pressure = 101.3KPa
V = volume = ?
n = number of moles = 4.5moles
T = Temperature = 273.15K
R = molar gas constant = 8.314J/mol.k
Rearranging the equation to make V the subject of the formula yields :
V = nRT/P
= ( 4.5 × 8.314 × 273.15) ÷ 101.3
= 10,219.361 ÷ 101.3 = 100.88L which is apprx 101L
Answer:
128gCaCl2 per 200gH2O
Explanation:
I might be wrong rlly srry if I am:/ Hope this helps doe have a wonderful day
