Brenda is correct!
Hope this helps!:)
5.7% KCl is 94.3 % water.
Therefore, for 1000 g of water the mass of KCl will be (1000× 5.7)/94.3 = 60.445 grams.
1 mole of KCl is equal to 74.55 g,
therefore, 60.445 g will be 60.445/74.55 = 0.8108 mole of KCl
Hence, 0.8108 moles of KCl should release twice that number of moles 1.6216 moles ions.
Having 1.6216 moles of KCl ions dissolved in 1000g of water, gives us 1.6216 molar if solution.
Using the freezing point depression constant of water.
dT = Kf (molarity)
dT = (1.86 C/ molar) (1.6216 m)
dT = 3.016 C drop in freezing point
Therefore, it should freeze at - 3.016 Celsius
CaC2 + 2 H2O = C2H2 + Ca(OH)2
moles CaC2 = 485 g/64.1 g/mol=7.57
moles H2O = 2 x 7.57 =15.1
mass H2O = 15.1 mol x 18.02 g/mol=272.8 g
moles C2H2 = 23.6 /26.038 g/mol=0.906
moles CaC2 = 0.906
mass CaC2 = 0.906 x 64.1 =58.1 g
moles Ca(OH)2 = 55.3 / 74.092 =0.760
moles H2O = 2 x 0.760 =1.52
mass water = 18.02 x 1.52 =27.4 g
100. g CCl4* (1 mol CCl4/ 153.8 g CCl4)* (6.02*10^23 CCl4 molecules/ 1 mol CCl4)= 3.91*10^23 CCl4 molecules.
(Note that the units cancel out so you get the answer)
Hope this helps~
0,26mol ------- 13g
1mol ------- x
x = (13g*1mol) / 0,26mol = 50g
<span>It could be Ti or V</span>