Answer:
B The water at the bottom of the lake will never freeze
Explanation:
Its simple
Answer:
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Explanation:
Data Given
M1 = 6.00 M
M2 = 2.5 M
V1 = 250 mL
V2 = ?
Solution:
As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.
Now
first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution
For this Purpose we use the following formula
M1V1=M2V2
Put values from given data in the formula
6 x V1 = 2.5 x 250
Rearrange the equation
V1 = 2.5 x 250 /6
V1 = 104 mL
So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide
But we have to prepare 250 mL of the solution.
so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.
in this question you have to tell about the amount of water that is 146 mL
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Answer:
pH 
Explanation:
For every mole of hydrochloric acid, one mole of hydronium ion is required. Thus, in order to neutralize 0.014 moles of HCL, 0.014 moles of hydronium is required.
![[H_3O^+] = [HCl] = 0.014](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%20%5BHCl%5D%20%3D%200.014)
pH ![= -log [H^+] = -log [H_3O^+]](https://tex.z-dn.net/?f=%3D%20-log%20%5BH%5E%2B%5D%20%3D%20-log%20%5BH_3O%5E%2B%5D)
Substituting the available values in above equation, we can say that the pH of the solution is equal to

pH 
pH of a
M HCL solution 
Explanation:
Given
The enthalpy of formation of RbF (s) is –557.7kJ/mol
The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol
The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)
= -583.8 - (-557.7) kJ/mol
= -26.1 kJ/mol
The enthalpy is negative which means that the temperature will rise when RbF is dissolved.
Answer:
3139542g
Explanation:
That's because if you talk about
4.5
moles of sodium fluoride, you would get a really absurd number of grams.