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.

Here's the balanced equation for given Double displacement reaction ~

The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is,
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in
of ethane gas
And, as we know that
1 mole of ethane molecule contains
molecules of ethane
2.869 moles of ethane molecule contains
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is,
molecule.
Answer:
Transition metals, alkali metals, alkaline earth metals Transition metals - Middle of the periodic chart, only average reactivity. alkali metals - As mentioned above, very reactive. Bad choice, going from lower reactivity to higher reactivity.
Hope this answer is right!