<span>(15.0 g) / (150.0 g) x (100 g) = 10.0 g/100 g H2O </span>
Answer:
See explanation
Explanation:
Molar mass of NaCl = 58.5 g
Number of moles contained in 10 g of NaCl = 10 g/58.5 g = 0.17 moles
If 1 mole of NaCl contains 6.02 * 10^23 atoms
0.17 moles of NaCl contains 0.17 * 6.02 * 10^23 atoms = 1.02 * 10^23 atoms
Molar mass of Fe II chloride = 126.751 g/mol
Number of moles = 10 g/126.751 g/mol = 0.0789 moles
Number of atoms = 0.0789 moles * 6.02 * 10^23 atoms = 4.7 * 10^22 atoms
Molar mass of Na = 23 g/mol
Number of moles = 10g/23 g/mol = 0.43 moles
Number of atoms = 0.43 moles * 6.02 * 10^23 atoms = 2.59 * 10^ 23 atoms
Answer:
i dont have a answer im just answering so you can give the person who answered brailiest :)
Explanation:
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
The answer is: the percent composition of chromium is 20.52%.
Ar(Cr) = 52; atomic mass of chromium.
Mr(BaCrO₄) = 253.3; molecular mass of barium chromate.
ω(Cr) = Ar(Cr) ÷ Mr(BaCrO₄) · 100%.
ω(Cr) = 52 ÷ 253.3 ·100%.
ω(Cr) = 20.52%; the percent composition of chromium in barium chromate.
