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2 years ago

Exactly 1 mol Hg2(NO3)2 contains how many moles of Hg , N , and O ?

Chemistry

1 answer:

marin [14]2 years ago

6 0

In exactly 1 mol Hg₂(NO₃)₂ , there are 2 mol Hg, 2 mol N and 6 mol O.

Since the molecular formula of Hg₂(NO₃)₂ shows that for every mole of the substance, we have 2 moles of Hg, 2 moles of N and 6 moles of O.

So, in exactly 1 mol Hg₂(NO₃)₂ , there are 2 mol Hg, 2 mol N and 6 mol O.

Learn more about number of moles here:

brainly.com/question/3935424

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Explanation:

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4 years ago

At what temperatures is a reaction that has a positive change in entropy

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Answer:

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3 years ago

Use the diagram to determine: How many atoms are on the Right side =___ How many atoms are on the Left side = ____ Is this a Bal

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4 years ago

Given the reaction: 4A1 + 302 - 2A1,O, How many grams of Al will be

dlinn [17]

Answer:

The answer to your question is 27 g of Al

Explanation:

Data

mass of Al = ?

moles of Al₂O₃ = 0.5

The correct formula for the product is Al₂O₃

Balanced chemical reaction

4Al + 3O₂ ⇒ 2Al₂O₃

Process

1.- Calculate the molar mass of the product

Al₂O₃ = (27 x 2) + (16 x 3)

= 54 + 48

= 102 g

2.- Convert the moles of Al₂O₃ to grams

102 g ---------------- 1 mol

x ---------------- 0.5 moles

x = (0.5 x 102) / 1

x = 51 g of Al₂O₃

3.- Use proportions to calculate the mass of Al

4(27) g of Al --------------- 2(102) g of Al₂O₃

x --------------- 51 g

x = (51 x 4(27)) / 2(102)

x = 5508 / 204

x = 27 g of Al

4 0

3 years ago

1. A given mass of air has a volume of 6.00 L at 80.0°C. At constant pressure, the temperature is

earnstyle [38]

3 L will be the final volume for the gas as per Charle's law.

Answer:

Explanation:

The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.

V∝ T

Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be

$\frac{V_{1} }{T_{1} }= \frac{V_{2} }{T_{2} }$

So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.

$\frac{6}{80}=\frac{V_{2} }{40}$

$V_{2}=\frac{6*40}{80}=3 L$

So, 3 L will be the final volume for the gas as per Charle's law.

4 0

3 years ago