Answer:
o.251 prduces 45.7L of oxogen
Explanation:
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<u>Answer:</u>
I think it's (C)
The products are suitable for making nuclear weapons.
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526 L O2 x 1 mol O2 / 22.4 L = 23.5 mol O2
<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:
Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL