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:
a. 52.8
Explanation:
To find the number of moles of HCl we use the relation M₁V₁=M₂V₂
where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.
M₁=7.91 M
M₂=2.13 M
V₁=?
V₂=196.1 mL
Replacing these values in the relationship.
M₁V₁=M₂V₂
7.91 M× V₁=2.13 M×196.1 mL
V₁=(2.13 M×196.1 mL)/7.91 M
=52.8 mL
I'm assuming false but really have no clue
Answer:
number of Al atom =54÷27=2atom of Al. number of o atom = 160÷16=10 atom of o . =214U MASS of ALO2.
B. The sand increases friction by increasing roughness.
