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3 years ago

Question 3

Chemistry

1 answer:

Dmitrij [34]3 years ago

4 0

Answer:

snow and very low temperatures are predicted for tomorrow.

Explanation:

snow and very low temperatures are associated with average weather condition of a place thus making it a statement about climate.

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mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

1.24 moles of magnesium arsenate are dissolved in 1.74 kg of solution. Calculate the molality of the solution.

aleksklad [387]

Answer:

Molality of the solution = 0.7294 M

Explanation:

Given:

Number of magnesium arsenate = 1.24 moles

Mass of solution = 1.74 kg

Find:

Molality of the solution

Computation:

Molality of the solution = Mole of solute / Mass of solution = 1.74 kg

Molality of the solution = 1.24 / 1.7

Molality of the solution = 0.7294 M

6 0

3 years ago

Zn + I2 ---> Znl2 Determine the theoretical yield of the product if 2g of Zn is used

egoroff_w [7]

Answer:

Mass = 9.58 g

Explanation:

Given data:

Mass of Zn = 2g

Theoretical yield of ZnI₂ = ?

Solution:

Chemical equation:

Zn + I₂ → ZnI₂

Number of moles of Zn:

Number of moles = mass/molar mass

Number of moles = 2g / 65.38 g/mol

Number of moles = 0.03 mol

Now we will compare the moles of Zn and ZnI₂.

Zn : ZnI₂

1 : 1

0.03 : 0.03

Mass of ZnI₂:

Mass = number of moles × molar mass

Mass = 0.03 mol × 319.22 g/mol

Mass = 9.58 g

4 0

3 years ago

How many significant figures does 0.005505 have

Viktor [21]

Answer:

it has six significant figures

8 0

3 years ago

To what temperature does a 250 ml cylinder containing 0.40 grams of helium gas need to be collected in order for the pressure to

VMariaS [17]

Answer:

T =76.13 K

Explanation:

Given data:

Temperature of gas = ?

Volume of gas = 250 mL(250/1000 = 0.25 L)

Mass of helium = 0.40 g

Pressure of gas = 253.25 kpa (253.25/101 = 2.5 atm)

Solution:

Formula:

PV = nRT

First of all we will determine the number of moles of helium.

Number of moles = mass/ molar mass

Number of moles = 0.40 g/ 4 g/mol

Number of moles = 0.1 mol

Now we will put the values.

R = general gas constant = 0.0821 atm.L/ mol.K

T = PV/nR

T =2.5 atm× 0.25 L /0.1 mol ×0.0821 atm.L/ mol.K

T = 0.625 /0.00821/K

T =76.13 K

6 0

3 years ago