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3 years ago

How many moles of KBr are present in 8.96 x 10^21

Chemistry

1 answer:

Gelneren [198K]3 years ago

6 0

Answer:

0.015 moles

Explanation:

One mole of a compound contains molecules equivalent to the Avogadro's constant, 6.022 × 10^23.

That is, 1 mole of a compound will have 6.022 × 10^23 molecules.

In our case, We are given 8.96 x 10^21 molecules of KBr

We need to find the number of moles in 8.96 x 10^21 molecules

1 mole of KBr = 6.022 × 10^23 molecules.

8.96 x 10^21 molecules = ?

Therefore;

(1 × 8.96 x 10^21 molecules ) ÷ 6.022 × 10^23 molecules.

= 1.488 × 10^-2 moles

= 0.01488 moles

= 0.015 moles

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Answer:

Explanation:

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3 years ago

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CO_2 sublimes readily at 25°C. Which properties are usually associated with a compound that undergoes this kind of change?

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Explanation:

Sublimation is defined as a process in which solid state of a substance directly changes into vapor or gaseous state without undergoing liquid phase.

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As this process does not cause any change in chemical composition of a substance. Hence, it is known as a physical process.

Similarly, when $CO_{2}$ sublimes readily at $25^{o}C$ . This shows change in physical state of carbon dioxide is taking place, i.e, from solid to gaseous phase.

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3 years ago

Please somebody give me the answers

RSB [31]

Answer:

1. 9.4 grams of methane produce 25.85 grams of CO2

2.Grams of water produced = 11.81 grams

3.Mass of Methane produced by 10.1 gram of O2 = 2.52 grams

4.Amount of methane consumed = 46.9 grams

5. Grams of Co2 produced = 8.32 grams

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

1 mole of CH4 = 16 gram

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams (1 mole of CO2 = 44 gram )

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

$1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)$

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

$\frac{44}{16}$ gram of CO2

9.4 gram of CH4 =

$\frac{44}{16}\times 9.4$ gram of CO2

= 25 .85 gram of CO2

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

$\frac{18}{32}$

21 gram of O2 =

$\frac{18}{32}\times 21$

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32) = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

$\frac{16}{64}$

10.1 gram of O =

$\frac{16}{64}\times 10.1$ of CH4

= 2.52 gram

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

$\frac{16}{44}$ gram of CH4

129 gram of CO2 =

$\frac{16}{44}\times 129$ gram of CH4

= 46.90 grams

2 mole of O2 produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

$\frac{44}{64}$ of CO2

12.1 gram of O2 produce=

$\frac{44}{64}\times 12.1$ of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

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3 years ago

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Answer:

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Explanation:

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Answer:

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Explanation:

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Solution:

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3 years ago