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3 years ago

How many moles are in 9.8 L of xenon gas?

Chemistry

1 answer:

mel-nik [20]3 years ago

4 0

Answer:

0.4375 moles

Explanation:

At standard temperature and pressure one mole of ideal gas occupies 22.4 L of volume.

Hence we can write following relation,

22.4 L volume is occupied by = 1 mole of Gas

So,

9.8 L of Volume will be occupied by = X mole of Gas

Solving for X,

X = 1 mole × 9.8 L / 22.4 L

X = 0.4375 moles

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C1 Progress quiz: Atomic structure 2 - test

myrzilka [38]

Answer:

0.11 nm

Explanation:

1.1 x 10-10 m

The goal is to convert the atomic radius from meters (m) to nanometres (nm). We do this by multiplying the value by 10^9.

This is given as;

$1.1 * 10^{-10} * 10^{9}\\1.1 * 10^{-10 + 9}\\1.1 * 10^{-1}\\0.11$

5 0

3 years ago

An iron bar of mass 841 g cools from 84°C to 7°C. Calculate the heat released (in kilojoules) by the metal.

Oduvanchick [21]

Answer:

28.75211 kj

Explanation:

Given data:

Mass of iron bar = 841 g

Initial temperature = 84°C

Final temperature = 7°C

Heat released = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

specific heat capacity of iron is 0.444 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 7°C - 84°C

ΔT = -77°C

By putting values,

Q = 841 g × 0.444 j/g.°C × -77°C

Q = 28752.11 j

In Kj:

28752.11 j × 1 kJ / 1000 J

28.75211 kj

8 0

3 years ago

The density of pure gold is 19.3g/cm3. What is the volume of 1.00 g of pure gold?

soldier1979 [14.2K]

Answer: Volume of 1g of pure gold $v=0.05181 \mathrm{cm}^{3}$

Given;

Density of a pure gold=19.3 $g / c m^{3}$

Mass of a pure gold =1g

To find:

Volume of 1g of pure gold

Solution:

According to the formula,

Density= Mass/Volume

$\rho=m / v$

Where $\rho$ =density of pure gold

m=mass of pure gold

v=volume of pure gold

From the above equation volume can be calculated as

$v=m / \rho$

Substitute the values of mass and density value in the above equation

$v=1 / 19.3$

$v=0.05181 \mathrm{cm}^{3}$

Result:

Thus the volume of 1g of pure gold is $v=0.05181 \mathrm{cm}^{3}$

8 0

3 years ago

Suppose you have an Avogadro's number of nitrogen atoms. How many grams does this represent? Enter your answer to one decimal pl

shusha [124]

Answer:

2.3 x 10-23 g.

Explanation:

The mass of a single atom is the mass number, 14, is the mass in grams of one mole of carbon.

One mole of Nitrogen atom is 6.022 x 1023 atoms (Avogadro's number). This can then used to convert a nitogen atom to grams by the ratio:

mass of 1 atom / 1 atom = mass of a mole of atoms / 6.022 x 10^23 atoms.

mass of 1 atom = mass of a mole of atoms / 6.022 x 1023

mass of 1 N atom = 14 / 6.022 x 10^23 N atoms

mass of 1 N atom = 2.325 x 10^-23 g

The mass of a single Nitrogen atom is 2.325 x 10-23 g.

3 0

3 years ago

You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene. Upon complete combustion in oxygen, you collect 0.204 g o

Elan Coil [88]

Answer:

The empirical formula is C7H5N3O6

Explanation:

Step 1: Data given

Mass of sample = 0.150 grams

Mass of CO2 = 0.204 grams

Molar mass CO2 = 44.01 g/mol

Mass of H2O = 0.030 grams

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 0.204 grams / 44.01 g/mol

Moles CO2 = 0.00464 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.00464 moles we have 0.00464 moles C

Step 4: Calculate mass C

Mass C = 0.00464 moles * 12.01 g/mol

Mass C = 0.0557 grams

Step 5: Calculate moles H2O

Moles H2O = 0.030 grams / 18.02 g/mol

Moles H2O = 0.00166 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00166 moles H2O we have 2* 0.00166 = 0.00332 moles H

Step 7: Calculate mass H

Mass H = 0.00332 moles * 1.01 g/mol

Mass H = 0.00335 grams

Step 8: Calculate mass N

Mass N = 0.185 * 0.150 grams

Mass N = 0.02775 grams

Step 9: Calculate moles N

Moles N = 0.02775 grams / 14.0 g/mol

Moles N = 0.00198 moles

Step 10: Calculate mass O

Mass O = 0.150 grams - 0.02775 - 0.00335 - 0.0557

Mass O = 0.0632 grams

Step 11: Calculate moles O

Moles O = 0.0632 grams / 16.0 g/mol

Moles O = 0.00395 moles

Step 11: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.00464 moles / 0.00198 moles =2.33

H: 0.00332 moles / 0.00198 moles = 1.66

N: 0.00198 moles / 0.00198 moles = 1

O: 0.00395 moles / 0.00198 moles = 2

For 1 mol N we have 2.33 moles C, 1.66 moles H and 2 moles O

For 3 moles N we have 7 moles C, 5 moles H and 6 moles O

The empirical formula is C7H5N3O6

5 0

3 years ago