Answer:
0.544 M
Explanation:
First find the moles in the final solution
0.8 mols/L *1.7L
1.36 mols
so there is 1.36 mols in 2.5L
concentration will be 1.36/2.5
0.544 M
Answer:
See the answer below
Explanation:
The best approach would be to <u>pour the liquid from the large reagent bottle into a small-size beaker or reagent bottle first</u>, before measuring the required quantity out into the reaction vessel. This is necessary in order to maintain safety in the laboratory.
Pouring the liquid directly from the large reagent bottle into the measuring cylinder or directly into the reaction bottle can compromise safety in the laboratory. The liquid might splash out and cause harm to the handler or create other harmful circumstances in the laboratory.
Answer:
The atomic mass of phosphorus is 29.864 amu.
Explanation:
Given data:
Atomic mass of phosphorus = ?
Percent abundance of P-29 = 35.5%
percent abundance of P-30 = 42.6%
Percent abundance of P-31 = 21.9%
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass / 100
Average atomic mass = (29×35.5)+(30×42.6) + (31×21.9) /100
Average atomic mass = 1029.5 + 1278 + 678.9/ 100
Average atomic mass = 2986.4 / 100
Average atomic mass = 29.864 amu.
The atomic mass of phosphorus is 29.864 amu.
Answer: The ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
of a solution is 
M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in ml
moles of 
= 
Now put all the given values in the formula of molality, we get
pH or pOH is the measure of acidity or alkalinity of a solution.
According to stoichiometry,
1 mole of gives 1 mole of 
Thus 
moles of gives =
moles of
Putting in the values:
![[H^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
![[0.01][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5B0.01%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
![[OH^-]=10^{-12}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-12%7D)
Thus the of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is M
