Answer: Thus concentration of 
in 
is 0.011 and in 
is 0.814
Explanation:
To calculate the concentration of , 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
We are given:
Putting values in above equation, we get:
The concentration in is 
Thus concentration of is 
and 
Answer : The change in entropy is 
Explanation :
Formula used :
where,
= change in entropy = ?
m = mass of water = 1.00 kg
= heat of vaporization of water = 
T = temperature = 
Now put all the given values in the above formula, we get:
Therefore, the change in entropy is
Rip King von like to send love to their family 63-63= von
Answer:
C 2
Explanation:
2Nal + Cl2 → 2NaCl + I2
This is balanced equation
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻
1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions
the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol
NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol
total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
