The mass of cobalt (III) needed is
m = 5.2 L (0.42 mol/L) ( 93 g/mol)
m = 97.65 g
The volume of nitric acid needed is
V = 5.2 L (0.42 mol/L) (3 mol / 1 mol) (1000 mL/1.6 mol)
V = 1968.75 mL
The moles of water produced is
n = 5.2 L (0.42 mol/L) (3 mol / 1 mol)
n = 3.15 moles
Answer:
15 moles of ammonium sulfate would be formed from 30 moles of ammonia.
Explanation:
Given data:
Number of moles of ammonium sulfate formed = ?
Number of moles of ammonia = 30.0 mol
Solution:
Chemical equation:
2NH₃ + H₂SO₄ → (NH₄)₂SO₄
Now we will compare the moles of ammonium sulfate with ammonia.
NH₃ : (NH₄)₂SO₄
2 : 1
30.0 : 1/2×30.0 = 15.0 mol
So 15 moles of ammonium sulfate would be formed from 30 moles of ammonia.
Answer:
TATGGCGTT
Explanation:
Complimentary base pairs:
A-T
C-G
Use the other letter for complimentary strands
Answer:
32.7
Explanation:
I just did it and got it right
