It would have to increase pressure... but I don’t see that option here..?
To prepare 350 mL of 0.100 M solution from a 1.50 M
solution, we simply have to use the formula:
M1 V1 = M2 V2
So from the formula, we will know how much volume of the
1.50 M we actually need.
1.50 M * V1 = 0.100 M * 350 mL
V1 = 23.33 mL
So we need 23.33 mL of the 1.50 M solution. We dilute it
with water to a volume of 350 mL. So water needed is:
350 mL – 23.33 mL = 326.67 mL water
Steps:
1. Take 23.33 mL of 1.50 M solution
<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M
solution</span>
Answer:
Volume of HCl require = 6 mL
Explanation:
Given data:
Volume of HCl require = ?
Molarity of HCl solution = 1.60 M
Volume of NaOH = 48.0 mL
Molarity of NaOH = 0.200 M
Solution:
Formula:
M₁V₁ = M₂V₂
By putting values,
1.60 M×V₁ = 0.200 M×48.0 mL
V₁ = 0.200 M×48.0 mL/1.60 M
V₁ = 9.6 M .mL /1.60 M
V₁ = 6 mL
Answer:
The classification is mentioned below for the particular topic.
Explanation:
- Whether we position 2 different beakers in such a single beaker through one clean edge of zinc-containing H₃Po₄ and another one with unflushed zinc.
- The zinc that was washed set to release hydrogen gas way quicker, unlike unventilated zinc.
⇒ 
- Since fresh zinc complicates the cycle since, as a comparison to polluted zinc, there was little contact with either the reaction.
C
0.70
I hope this is help, I’m so so sorry if I’m incorrect
