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Alenkinab [10]
3 years ago
13

Copper atoms are heavier than magnesium atoms. So, although each atom of magnesium can produce one atom of copper, the masses wo

n’t be the same. The ratio of the atomic weight of copper to the atomic weight of magnesium is about 2.61. Given this ratio and the initial mass of the magnesium strip (0.38 g) measured in task 1, part A, calculate the mass of copper that can be produced. Recall: Mg + Cu(NO3)2 → Cu + Mg(NO3)2
Chemistry
1 answer:
Ilya [14]3 years ago
8 0

Answer:

m_{Cu}=0.99gCu

Explanation:

Hello!

In this case, considering the given chemical reaction and the mass of the magnesium strip, following the indications of the atomic weight ratio (2.61 g Cu/1 g Mg), and keeping in mind the 1:1 mole ratio one could compute the produced mass of copper as shown below:

m_{Cu}=0.38gMg*\frac{2.61gCu}{1gMg} \\\\m_{Cu}=0.99gCu

Best regards!

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The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What
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Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

=  Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

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The heat of vaporization at 75° C = 900 J/mol

8 0
4 years ago
For the complete combustion of 47 g of gasoline (octane, C8H18) , the mass of oxygen consumed is
aleksklad [387]

Answer: d) 164.9 g

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To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}     \text{Moles of octane}=\frac{47g}{114g/mol}=0.412moles

The balanced chemical reaction is:

2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)  

According to stoichiometry :

2 moles of C_8H_{18} require = 25  moles of O_2

Thus 0.412 moles of C_8H_{18} will require=\frac{25}{2}\times 0.412=5.15moles  of O_2

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3 years ago
What is the difference expressed in kilograms between the mass of proton and the mass of electon?
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4 0
4 years ago
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how many grams of ammonium carbonate are needed to decompose in order to produce 6.52 g of carbon dioxide?
defon

Answer:

<u>= 14.24g of </u>(NH4)_{2}CO_{3}<u> is required.</u>

Explanation:

Reaction equation:

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