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Free_Kalibri [48]
4 years ago
7

Can someone help me check my answers for the mole questions?

Chemistry
1 answer:
Sliva [168]4 years ago
3 0

Answer:

Explanation:

Given data:

Number of particles of phosphorous = 1.2 × 10²⁵

Number of moles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

Number of particles of phosphorous = 1.2 × 10²⁵ particles

1 mole = 6.022 × 10²³ particles

1.2 × 10²⁵ particles × 1 mole /  6.022 × 10²³ particles

0.1993  × 10² mol

19.93 mole

Q 3:

Given data:

Number of moles of calcium = 0.630 mol

Number of particles = ?

Solution:

The number 6.022 × 10²³ is called Avogadro number.

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ particles of hydrogen

For 0.63 moles:

1 mole = 6.022 × 10²³ particles

0.63 mol × 6.022 × 10²³ particles  / 1 mol

3.79 × 10²³ particles

Q 4:

Given data:

Number of moles = 4 mol

Number of particles = ?

Solution:

The number 6.022 × 10²³ is called Avogadro number.

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ particles of hydrogen

For 4 moles:

1 mole = 6.022 × 10²³ particles

4 mol × 6.022 × 10²³ particles  / 1 mol

24 × 10²³ particles

2.4  × 10²⁴ particles

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Identify the type of reaction shown by this chemical equation:<br>2AI + 6HCI - 2AlCl3 + 3H2​
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Single displacement reaction

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3 years ago
A compression, at a constant pressure of 140 kPa, is performed on 4.0 moles of an ideal monatomic gas (Cv = 3/2 R). The compress
kobusy [5.1K]

<u>Answer:</u> The change in internal energy of the gas is 29.414 kJ.

<u>Explanation:</u>

To calculate the temperature of the gas at different volumes, we use ideal gas equation:

PV=nRT

  • When volume = 0.26m^3

We are given:

Conversion used:  1m^3=1000L

P=140kPa\\V=0.23m^3=260L\\n=4.0mol\\R=8.31\text{L kPa }mol^{-1}K^{-1}

Putting values in above equation:

140kPa\times 260L=4mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T_i\\\\T_i=1095.06K

  • When volume = 0.12m^3

We are given:

P=140kPa\\V=0.12m^3=120L\\n=4.0mol\\R=8.31\text{L kPa }mol^{-1}K^{-1}

Putting values in above equation:

140kPa\times 120L=4mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T_f\\\\T_f=505.41K

  • To calculate the change in internal energy, we use the equation:

\Delta U=nC_v\Delta T=nC_v(T_f-T_i)

where,

\Delta U = change in internal energy = ?

n = number of moles = 4.0 mol

C_v = heat capacity at constant volume = \frac{3}{2}R

T_f = final temperature = 1095.06 K

T_i = initial temperature = 505.41 K

Putting values in above equation, we get:

\Delta U=4\times \frac{3}{2}\times 8.314J/K.mol\times (505.41-1095.06)\\\\\Delta U=29414.1J

Converting this value in kilojoules, we use the conversion factor:

1 kJ = 1000 J

So, 29414.1J=\frac{1kJ}{1000J}\times 29414.1J=29.414kJ

Hence, the change in internal energy of the gas is 29.414 kJ.

6 0
4 years ago
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Answer:

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Hope this is one of the answer choices!

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