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2 years ago

If 725 ml of N² gas is at 1 atm, what volume will it have at 1.4 atm?

Chemistry

2 answers:

nasty-shy [4]2 years ago

8 0

Answer:

517.86 mL

Explanation:

Boyle's Law Formula

P₁V₁ = P₂V₂

Here :

V₁ = 725 mL
P₁ = 1 atm
P₂ = 1.4 atm

Solving

V₂ = P₁V₁/P₂
V₂ = 1 × 725 / 1.4
V₂ = 517.86 mL

alexdok [17]2 years ago

7 0

Answer:

518 mL

Explanation:

We can solve this using Boyle's Law Formula

P1V1 = P2V2

where p1 = initial pressure, p2 = final pressure, v1 = initial volume and v2 = final volume

here , the initial pressure is 1 atm and the initial volume is 725mL

we are given the final pressure 1.4 and we need to find the final volume

so we have p1v1 = p2v2

==> plug in p1 = 1 , v1 = 725 mL and p2 = 1.4

(1)(725) = (1.4)v2

==> multiply 1 and 725

725 = (1.4)(v2)

==> divide both sides by 1.4

v2 = 518

N2 would have a volume of 518mL at 1.4atm

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What is the mass of 8.03 mole of NH3? (NH3 is 17g/mol)

lana [24]

Answer:

the mass of 8.03 mole of NH3 is 136.51 g

Explanation:

The computation of the mass is shown below:

As we know that

Mass = number of moles × molar mass

= 8.03 mol × 17 g/mol

= 136.51 g

Hence, the mass of 8.03 mole of NH3 is 136.51 g

We simply multiplied the number of moles with the molar mass so that the mass could come

3 0

2 years ago

Consider the balanced equation for the following reaction:

Zanzabum

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $\frac{4}{7}\times 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

7 0

2 years ago

A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder

Mazyrski [523]

Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

$PV_1=nRT$

where,

P = pressure of gas = 1 atm

$V_1$ = volume of gas = ?

T = temperature of gas = $270^oC=273+270=543K$

n = number of moles of gas = 0.205 mol

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

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K$

$V_1=9.12L$

Now we have to calculate the volume at 24°C.

$PV_2=nRT$

where,

P = pressure of gas = 1 atm

$V_2$ = volume of gas = ?

T = temperature of gas = $24^oC=273+24=297K$

n = number of moles of gas = 0.205 mol

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

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K$

$V_2=4.99L$

Now we have to calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 9.12 L

$V_2$ = final volume = 4.99 L

Now put all the given values in the above formula, we get:

$w=-p(V_2-V_1)$

$w=-(1atm)\times (4.99-9.12)L$

$w=4.13L.artm=4.13\times 101.3J=418.4J$

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

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2 years ago

Express the frequency in inverse seconds of n=4 to n=3

HACTEHA [7]

Answer:

1.60×10¹⁴ s⁻¹

Explanation:

When an electron jumps from one energy level to a lower energy level some energy is released in the form of a photon.

4 0

2 years ago

Question 8 (9 points) Match the lab equipment with its purpose

Deffense [45]

Answer:

$\sf \boxed{4} \mapsto Pipet$

$\sf\boxed{7}\mapsto Test \:tube \: rack$

$\sf\boxed{3}\mapsto Test\: table$

$\sf\boxed{5}\mapsto Scoopula$

$\sf\boxed{1}\mapsto Graduated\: cylinder$

$\sf\boxed{9}\mapsto Bunsen \:burner$

$\sf\boxed{2}\mapsto Beaker$

$\sf \boxed{8}\mapsto Spot\: plate$

$\sf\boxed{6}\mapsto Goggles$

Explanation:

Pipet is used to dispense a very small amount of liquid.

Test tube rack is used to hold multiple test tubes at the same time.

Test Table is used to view chemical reactions or hold or heat small amounts of substance.

Scoopula is used to dispense chemicals from a larger container.

Graduated cylinder is used to measure volume very precisely.

Bunsen burner is used to heat objects.

Beaker is used to transport heat or store substance.

Spot plate is used to observe the color changes of small quantities of a reacting mixture.

Goggles are used to protect the eyes from flying objects or chemical splashes.

_____________________________________________________

7 0

2 years ago

If 725 ml of N² gas is at 1 atm, what volume will it have at 1.4 atm?​

If 725 ml of N² gas is at 1 atm, what volume will it have at 1.4 atm?