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:
the results of an investigation thank me later
Explanation:
Answer:
The answer to your question is V = 0.108 L or 108 ml
Explanation:
Data
Volume = ?
mass = 0.405 g
Temperature = 273°K
Pressure = 1 atm
Process
1.- Convert mass of Kr to moles
83.8 g of Kr -------------------- 1 mol
0.405 g ------------------- x
x = (0.405 x 1) / 83.8
x = 0.0048 moles
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for V
V = nRT / P
- Substitution
V = (0.0048)(0.082)(273) / 1
- Simplification
V = 0.108 / 1
- Result
V = 0.108 L
The answer is b but you want to think of why it is moving fast because of the temp indication
