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4 years ago

2Na(s) + Cl2(g) ------ 2NaCl(s)

Chemistry

1 answer:

Dovator [93]4 years ago

6 0

Answer:

1. 7.256g of NaCl

2. 47.33g of Cl2

Explanation:

2 moles of Na reacts to produce 2 moles of NaCl

8 moles of Na will still produce 8 moles of NaCl

Mass of NaCl = molar mass of Nacl/moles of Nacl

=58.5/8

=7.256g of NaCl

From the equation, 2 moles of Na reacts with 1 mole of Cl2

3/2 moles of Cl2 will react with 3 moles of Na

Mass of Cl2 = 71/1.5

=47.33g of Cl2

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Some areas of the Earth receive more solar radiation than others. Which of the following results from the Sun's uneven heating o

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Further Explanations:

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Answer Details :

Grade: High School

Subject: Geography

Chapter: Climate

Keywords:

Earth, solar radiation, orbicular shape, June, sun, Tropic of cancer, Tropic of Capricorn,Climate,climatological variables, climate temperature, moistness, atmospheric pressure, current of air, and precipitation, daily temperature ,rainfall activities

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4 years ago

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Explanation:

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For every 1.0 mole of glycine in the sample, how many molecules of methionine are present? (for help performing calculations wit

krok68 [10]

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Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

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ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

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Nimfa-mama [501]

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