A second order reaction varies with the square of the concentration of the reactant. Therefore, halving the concentration will reduce the rate of reaction by a factor of 4.
The answer is E.
Explanation:
option option B is the correct answer of given statement helium-4(He)=2
Answer:
The new concentration is 0.125 M.
Explanation:
Given data:
Initial volume V₁ = 125.0 mL
Initial molarity M₁ = 0.150 M
New volume V₂ = 25 mL +125 mL = 150 mL
New concentration M₂ = ?
Solution:
M₁V₁ = M₂V₂
0.150 M × 125 mL = M₂ × 150 mL
M₂ = 0.150 M × 125 mL / 150mL
M₂ = 18.75 M.mL/150 mL
M₂ = 0.125 M
The new concentration is 0.125 M.
Thallium has got 81 protons
<u>Have a nice days.......</u>
OK, so to answer this question, you will simply use the molality equation which is as follows:
<span>M1V1 = M2V2
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml
</span>plug in the givens in the above equation:
<span>2 x V1 = 0.4 x 100
</span>therefore:
V1 = 20 ml
Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.
