SrSo4 = Sr(2+) + SO4(2-)
Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).
When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4
Answer:
see explanation for answer
Explanation:
salivary gland: 1
stomach: 2
small intestine: 6
liver: 4
gallbladder: 5
large intestine: 3
The answers correspond with the numbers on the text boxes, so you would drag number 1 to the salivary gland and so on.
Answer: -2m/s2
Explanation:
Using the following equation ; acceleration = Change in velocity / time
i.e a = v - u / t
where 'a' = acceleration
v = final velocity
u = initial velocity
t = time
Therefore; from the graph we have acceleration to be, 0 - 6m/s / 3s = -2m/s2
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)
