Answer:
2266g
Explanation:
mass = no.of molecules /6.o23*1o(23) * molar mass
molar mass of co2= 44g /mol
1.5 .10^25/6.023 .10^23 =51.5 moles of co2
51.5 .44g/mol =2266 g
The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
1)Reactants
2)Light
3)An item that can increase reaction rates
4)Reactants must collide with each other
Less molecules lower the chance for collisions
The more collisions there are the higher the reaction rate
