Answer:
0.005 M
Explanation:
Given data:
volume of sample solution ( volume of D ) = 5.0 mL
volume of added stock solution ( V1 ) = 5.0 mL
concentration of added stock solution ( N1 ) = 0.02 M
Total volume of concentration ( V2 )= 10 mL = ( 5.0 mL + 5.0mL)
concentration of Total volume of sample ( C2 ) = 0.01
N2 = ( N1V1 ) / V2
= ( 0.02 * 5 ) / 10 = 0.01 m
absorbance of sample solution ( A1 ) = 0.10
absorbance of additional sample solution ( A2 ) = 0.20
attached below is the remaining part of the detailed solution
Answer:
5.4 M.
Explanation:
- At complete neutralization: It is known that the no. of millimoles of acid equal that of the base.
<em>(MV)acid = (MV)NaOH</em>
M of acid = ??? M, V of acid = 35.0 mL.
M of NaOH = 3.0 M, V of NaOH = 63.0 mL.
∴ M of acid = (MV)NaOH / (V)acid = (3.0 M)(63.0 mL)/(35.0 mL) = 5.4 M.
Answer:
1s22s22p63s1 is the electronic configuration of sodium.
Explanation:
If 1mole ------------- is ---------------- 6.02*10²³
than 0.25mole ----- is ---------------- x
x = [0.25mole*6.02*10²³]/1mole = <u>1,505*10²³</u>