0.25 L Barium nitrate solution contains 0.02 mol nitrate ions
<h3>Further explanation</h3>
Given
0.04 M Barium nitrate, Ba(NO₃)₂(aq)
Required
The volume of Barium nitrate
Solution
Ionization of Barium nitrate in 1 L solution :
Ba(NO₃)₂ ⇒ Ba²⁺ + 2NO₃²⁻
mol :
0.04 0.04 0.08
There is 0.08 mol of nitrate ions in 1 L solution
So for 0.02 mol nitrate ions :
= 0.02/0.08 x 1 L
= 0.25 L
Answer:
0.4 moles of water produced by 6.25 g of oxygen.
Explanation:
Given data:
Mass of oxygen = 6.25 g
Moles of water produced = ?
Solution:
Chemical equation;
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 6.35 g/ 32 g/mol
Number of moles = 0.2 mol
Now we will compare the moles of oxygen with water:
O₂ : H₂O
1 : 2
0.2 : 2×0.2 = 0.4 mol
0.4 moles of water produced by 6.25 g of oxygen.
Answer:
7 hours
Explanation:
627/96=7
7
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