Answer:
Explanation:
So, just find 55 on the solubility side and with your finger just move to the right untill you touch the line you should get in between 46-48. I would go more for 48.
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True. Any amount of oil can contaminate water, no matter how much water there is or how much oil is added.
Answer:
has greater impact on the freezing point depression of ice.
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor
= freezing point constant
m= molality
a) i = 4 for
as it dissociates to give 4 ions in water.
![CrF_3\rightarrow Cr^{3+}+3F^-](https://tex.z-dn.net/?f=CrF_3%5Crightarrow%20Cr%5E%7B3%2B%7D%2B3F%5E-)
b) i = 3 for
as it dissociates to give 3 ions in water.
![CaF_2\rightarrow Ca^{2+}+2F^-](https://tex.z-dn.net/?f=CaF_2%5Crightarrow%20Ca%5E%7B2%2B%7D%2B2F%5E-)
As the vant hoff factor is higher for
, it has greater impact on the freezing point depression of ice.
It is nitric acid, <span>Nitric acid, also known as aqua fortis and spirit of niter, is a highly corrosive mineral acid. The pure compound is colorless, but older samples tend to acquire a yellow cast due to decomposition into oxides of nitrogen and water.</span><span>
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Answer:
B = 0.10 M of AlCl₃
Explanation:
A = 0.10 M NaCl
NaCl → Na⁺ + Cl⁻
sodium chloride dissociate into sodium and chlorine ions. The one mole of NaCl produced on mole of Na⁺ and one mole of Cl⁻, so concentration of chloride ions will be 0.10 M.
B = 0.10 M of AlCl₃
AlCl₃ → Al³⁺ + 3Cl⁻
AlCl₃ dissociate into aluminium cation and chlorine anion. one mole of AlCl₃ dissociate into one mole of Al³⁺ and three mole of Cl⁻.
so concentration of chloride ion in 0.10 M solution is
0.10 × 3 = 0.3 M
C = 0.10 M MgCl₂
MgCl₂ → Mg²⁺ + 2Cl⁻
MgCl₂ dissociate into one mole of Mg²⁺ and two mole of Cl⁻.
so concentration of chloride ions in 0.10 M solution of MgCl₂ is,
2× 0.10 M = 0.20 M
d = 0.05 M CaCl₂
CaCl₂ → Ca²⁺ + 2Cl⁻
CaCl₂ dissociate into one mole of Ca²⁺ and two mole of Cl⁻, So concentration of chloride ions in 0.05 M CaCl₂ is
2× 0.05 = 0.1 M