1. D
2. DR
3. SR
4. DR
5. C
6. SR
7. S
8. D
10. C
We know,
AgNO3 + NaCl ⇒ NaNO3 + AgCl(s)
The moles of Na+ present:
0.5 L * 0.001 mol/L
= 5 x 10⁻⁴ mol
Moles of Ag+ present:
0.5 * 0.02
= 0.01 mol
The limiting reactant is Na
Therefore, the moles of Ag reacted:
5 x 10⁻⁴
AgCl is insoluble in water; therefore, the AgCl formed will precipitate
Answer:
w w w . g o o g l e . c o m
Explanation:
Answer: Thus molarity of 
is 1.35 M
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
Thus molarity of is 1.35 M
