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klio [65]
3 years ago
10

How many moles of Iron Fe, are in 0.054g of iron?​

Chemistry
1 answer:
Ede4ka [16]3 years ago
8 0

Answer:

9.643×10^-4 mole is your answer

Explanation:

56 g of iron =1 mole

0.054 g of iron =1×0.054/56=9.643×10^-4 mole

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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
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Answer:

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Explanation:

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T_{initial} = initial temperature = 28.9^oC

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q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

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q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

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