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Sophie [7]
3 years ago
9

A chemist measures the amount of bromine liquid produced during an experiment. She finds that 14.4 g of bromine liquid is produc

ed. Calculate the number of moles of bromine liquid produced.
Round your answer to 3 significant digit
Chemistry
1 answer:
emmasim [6.3K]3 years ago
5 0

Answer:

\boxed{\text{0.0901 mol}}

Explanation:

\text{Moles} = \text{14.4 g } \times \dfrac{\text{1 mol}}{\text{159.81 g}} = \text{0.0901 mol}\\\\\text{ The sample contains $\boxed{\textbf{0.0901 mol}}$ of bromine}

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If the molar heat of fusion of water is 6.01
Zolol [24]

Answer : The heat energy required to melt 2 kg of ice was, 667.7 kJ

Explanation :

First we have to calculate the moles of ice.

\text{Moles of ice}=\frac{\text{Given mass ice}}{\text{Molar mass ice}}=\frac{2kg}{18g/mol}=\frac{2000g}{18g/mol}=111.1mol

Now we have to calculate the heat energy.

As, heat energy required to melt 1 mole of ice = 6.01 kJ

So, heat energy required to melt 111.1 mole of ice = 111.1 × 6.01 kJ

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6 0
4 years ago
Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat
Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values ​​of the atomic mass of each element that form the compounds:

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By stoichiometry, they react and occur in moles:

  • NH₃: 4 moles
  • O₂: 5 moles
  • NO: 4 moles
  • H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

  • NH₃: 4 moles*17 g/mol= 68 g
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  • NO: 4 moles*30 g/mol= 120 g
  • H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }

mass of O₂≅17.65 g

<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>

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