<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
Lettuce
Vegetable.
Leaf lettuce.
Head lettuce.
Romaine lettuce.
Asparagus lettuce
40.6 kJ of heat energy had been emitted.
CO(g) + 2H2(g) CH3OH(l)CO volume, V (CO), equals 15 L or 0.015 m3.
Temperature = 85 0C = 85 + 273 = 358 K Pressure = 112 kPa = 112,000 PaPV = nRT n= 112000 0.015 / 8.314 358 n(CO) = 0.56 moles,
according to the ideal gas law.H2 volume is 14.4 L or 0.0144 m3
T = 750C + 273 K = 348 K n(H2) = 99191.84 0.0144 m3 / 8.314 348 K = 0.49 moles of H2 Pressure = 744 torr = 99191.84 Pa
Hydrogen is the limiting reagent, according to the calculation above.CH3OH = H2 = 0.49/2 = 0.245 m-238.6 (-110.5) = -128.1 kJ/mol for H(rxn) = H(f) (CH3OH) - H (rxn)
We must now multiply H(rxn) by the number of moles of methanol.
E = H(rxn) n(CH3OH) = 128.1 0.245 = 40.6 kJ.
Learn more about Ideal gas law here-
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