Answer:
0.480 grams
Explanation:
Li₃N(s) + 3D₂O (L) --------------------------> ND₃(g) + 3LiOD (aq)
1 : 3 : 1 : 3
Number of moles (n) = Mass in gram/ Molar Mass
Mass of ND₃ = 160 mg
= 0.16 g
Molar mass of ND₃= [14 + (3 x 2.014 )]
= 14 + 6.042
= 20.042 g/mol
Number of moles of ND₃ = 0.16/20.042
= 0.007983 moles
From the reaction equation, the mole ratio between Heavy water (D₂O ) and ND₃ is 3: 1.
This implies that the number of moles of Heavy water (D₂O ) required
= 3 x 0.007983 moles
= 0.023949 moles
Molar mass of Heavy water (D₂O )= [(2.014 x 2) + 16]
= 20.028 g/mol
Mass in grams of Heavy water (D₂O )= Number of moles x Molar mass
= 0.023949 x 20.028
= 0.4797 grams
≈ 0.480 grams
Answer:
pH of buffer =4.75
Explanation:
The pH of buffer solution is calculated using Henderson Hassalbalch's equation:
![pH=pKa+log[\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5B%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
Given:
pKa = 3.75
concentration of acid = concentration of formic acid = 1 M
concentration of salt = concentration of sodium formate = 10 M
![pH=3.75+log[\frac{10}{1}]=3.75+1=4.75](https://tex.z-dn.net/?f=pH%3D3.75%2Blog%5B%5Cfrac%7B10%7D%7B1%7D%5D%3D3.75%2B1%3D4.75)
pH of buffer =4.75
Displaced volume:
final volume - initial volume
1 mL = 1 cm³
38.5 mL - 35.0 mL = 3.5 cm³
hope this helps!
Answer: 2 Na (s) + Cl(g) -> 2 NaCl (s)
Explanation:
Both products will start to cancel the acidity and how strong the base is if they are mixed. If the acid is stronger than the base then it will be an acidic product and visa versa if the base is stronger than the acid.
