Answer:
The balanced equation is:
2 HNO3 + Mg ---> Mg(NO3)2 + H2
From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg
Moles of Mg:
Molar mass of Mg = 24 g/mol
Moles = Given mass / Molar Mass
Moles of Mg = 4.47 / 24 = 0.18 moles (approx)
Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed
Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3
Unreacted moles of HNO3 = Total Moles - Moles consumed
Unreacted moles of HNO3 = 0.64 moles (approx)
Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate
From the given options, we can see that 0.632 moles is the closest value to our answer
Therefore, 0.632 moles will remain after the reaction
D. is the answer
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Answer:
Answer choice B
Explanation:
Since you do not know the volume of the liquid in each beaker, the one in the smaller beaker could have more substance and therefore more thermal energy. If they had the same amount of substance, then the more voluminous one would radiate faster. However, since you do not know this, there is no way to tell. PM me if you have more questions. Hope this helps!
Answer:
The enthalpy of atomization is the enthalpy change that accompanies the total separation of all atoms in a chemical substance. This is often represented by the symbol ΔₐₜH or ΔHₐₜ. All bonds in the compound are broken in atomization and none are formed, so enthalpies of atomization are always positive.
Explanation:
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12 moles of water H₂O are produced from the combustion of pentane.
Explanation:
We have the following combustion of pentane (C₅H₁₂):
C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O
Knowing the chemical reaction we devise the following reasoning:
if 1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O
then 2 moles of pentane C₅H₁₂ produces X moles of water H₂O
X = (2 × 6) / 1 = 12 moles of water H₂O
Learn more about:
combustion of organic compounds
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