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3 years ago

15

Which of Graphs 1 correctly represents the relationship between the pressure and Kelvin temperature of a gas?

1 answer:

aliya0001 [1]3 years ago

6 0

Answer:

B

Explanation:

Pressure is directly proportional to temperature

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In sodium chloride the ions are lined up in a __ patterns

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The kinetic molecular theory of matter states that: Matter is made up of particles that are constantly moving. ... Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy. The temperature of a substance is a measure of the average kinetic energy of the particles.

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Nickel carbonyl, Ni(CO)4 is one of the most toxic substances known. The present maximum allowable concentration in laboratory ai

vodomira [7]

<u>Answer:</u> The mass of $Ni(CO)_4$ allowable in the laboratory is 4599.5 grams

<u>Explanation:</u>

To calculate the volume of cuboid, we use the equation:

$V=l\times b\times h$

where,

V = volume of cuboid

l = length of cuboid = 14 ft

b = breadth of cuboid = 22 ft

h = height of cuboid = 9 ft

Putting values in above equation, we get:

$V=14\times 22\times 9=2772ft^3=78503.04L$ (Conversion factor: $1ft^3=28.32L$

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

$PV=nRT$

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 78503.04 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

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

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$1.00atm\times 78503.04L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 296K\\\\n=\frac{1.00\times 78503.04}{0.0821\times 296}=3230.4mol$

Applying unitary method:

For every 109 moles of gas, the moles of $Ni(CO)_4$ present are 1 moles

So, for 3230.4 moles of gas, the moles of $Ni(CO)_4$ present will be = $\frac{1}{109}\times 3230.4=26.94mol$

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

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

Moles of $Ni(CO)_4$ = 26.94 moles

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$26.94mol=\frac{\text{Mass of }Ni(CO)_4}{170.73g/mol}\\\\\text{Mass of }Ni(CO)_4=(26.94mol\times 170.73g/mol)=4599.5g$

Hence, the mass of $Ni(CO)_4$ allowable in the laboratory is 4599.5 grams

3 0

3 years ago