Answer : The mass of sodium fluoride added should be 0.105 grams.
Explanation : Given,
The dissociation constant for HF = 
Concentration of HF (weak acid)= 0.0310 M
First we have to calculate the value of
.
The expression used for the calculation of
is,

Now put the value of
in this expression, we get:



Now we have to calculate the concentration of NaF.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[NaF]}{[HF]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BNaF%5D%7D%7B%5BHF%5D%7D)
Now put all the given values in this expression, we get:
![2.60=3.17+\log (\frac{[NaF]}{0.0310})](https://tex.z-dn.net/?f=2.60%3D3.17%2B%5Clog%20%28%5Cfrac%7B%5BNaF%5D%7D%7B0.0310%7D%29)
![[NaF]=0.00834M](https://tex.z-dn.net/?f=%5BNaF%5D%3D0.00834M)
Now we have to calculate the moles of NaF.

Now we have to calculate the mass of NaF.

Therefore, the mass of sodium fluoride added should be 0.105 grams.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
Can you give me the statements please?
Answer:
option D= Gold (I) nitride
Explanation:
The name of the given compound is gold(I) nitride.
Molar mass can be determine by following way:
molar mass Au3N = (molar mass of gold × 3) + (molar mass of nitrogen)
molar mass Au3N = (196.97 × 3 ) + ( 14 )
molar mass of Au3N = 590.91 g/mol + 14 g/mol
molar mass of Au3N = 604.91 g/mol
The nitrogen has valency of -3 so three Au(+1) will require while the valency of Au is (1+) one nitrogen will require to make the compound overall neutral.
Au3N
3(1+) + (-3) = 0
+3 - 3 = 0
0 = 0
The overall charge is 0, the compound will be neutral.
This is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions