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andrew-mc [135]
3 years ago
13

How many grams are in 3.1 x 10^11 atoms of chlorine

Chemistry
1 answer:
Kamila [148]3 years ago
3 0
6,02*10^{23} \ \ \ \ \ \rightarrow \ \ \ 35,5g\\
3,1*10^{11} \ \ \ \ \ \ \rightarrow \ \ \ x\\\\
x=\frac{3,1*10^{11}*35,5g}{6,02*10^{23}}\approx 18,3*10^{-2}g=0,183g
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Given:
bulgar [2K]

Answer:

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Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

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= 3.73125 moles x  -890 kJ/mol

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What is20 grams of HNO^3
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63.01284*20=1260.568

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What is the result of multiplying (2.5 × 1010) × (2.0 × 10-7)?
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That would be A.

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When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s) c) How many moles of the e
sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

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b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

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As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

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Excess Cl₂ = 212.7 - 159.64

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d) Calculate the moles of Cl

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                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

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