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4 years ago

20 PTS!!! What is the volume of a gas at 90.0°C, if it occupies 1.41 L at 170°C?

Chemistry

2 answers:

Fofino [41]4 years ago

5 0

Answer:

The new volume of the gas is 1.72L

Explanation:

V1 = 1.41L

V2 = ?

T1 = 90°C = (90+273.15)K = 363.15K

T2 = 170°C =(170+273.15)K = 443.15K

To find the new volume V2 of the gas, we apply Charles law which states that the volume of a fixed mass of gas is directly proportional to its temperature provided that pressure remains constant.

V = KT, K = V1 / T1 = V2 /T2 = V3 /T3.......= Vn / Tn

V1 / T1 = V2 / T2

Solve for V2,

V2 = (V1 * T2) / T1

V2 = (1.41 * 443.15) / 363.15

V2 = 1.72L

The new volume of the gas is 1.72L

Margarita [4]4 years ago

3 0

Answer:

1.72 L

Explanation:

Initial Temperature of the gas T =90°+273= 363 K

Initial Volume of the gas V1= 1.41 L

Final temperature of the gas T2= 170°c +273= 443 K

Final volume of the gas V2= ????

Using Charles law;

V1/T1 = V2/T2

V1T2= V2T1

V2= V1T2/T1

V2= 1.41×443/363

V2= 1.72 L

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Chloral hydrate (C2H3Cl3O2) is a drug formerly used as a sedative and hypnotic.

Dmitrij [34]

Answer :

(a) The molar mass of $C_2H_3Cl_3O_2$ is, 165.5 g/mole

(b) The moles of $C_2H_3Cl_3O_2$ is, 3.02 moles

(c) The mass in grams of $2.0\times 10^{-2}$ mole chloral hydrate is, 3.31 g

(d) The number of chlorine atoms in 5.0 g chloral hydrate is, $5.4\times 10^{22}$

(e) The mass of chloral hydrate will be, 1.55 g

(f) The mass of exactly 500 molecules of chloral hydrate is, $1.99\times 10^{23}$

Explanation :

(a) To calculate the molar mass of chloral hydrate.

The formula of chloral hydrate is, $C_2H_3Cl_3O_2$

Atomic mass of carbon = 12 g/mole

Atomic mass of hydrogen = 1 g/mole

Atomic mass of oxygen = 16 g/mole

Atomic mass of chlorine = 35.5 g/mole

Now we have to determine the molar mass of chloral hydrate.

$\text{Molar mass of }C_2H_3Cl_3O_2$ =2(12g/mole)+3(1g/mole)+3(35.5g/mole)+2(16g/mole)=165.5g/mole[/tex]

The molar mass of $C_2H_3Cl_3O_2$ is, 165.5 g/mole

(b) Now we have to determine the moles of $C_2H_3Cl_3O_2$ .

$\text{Moles of }C_2H_3Cl_3O_2=\frac{\text{Mass of }C_2H_3Cl_3O_2}{\text{Molar mass of }C_2H_3Cl_3O_2}=\frac{500.0g}{165.5g/mole}=3.02moles$

The moles of $C_2H_3Cl_3O_2$ is, 3.02 moles

(c) Now we have to determine the mass in grams of $2.0\times 10^{-2}$ mole chloral hydrate.

$\text{Mass of }C_2H_3Cl_3O_2=\text{Moles of }C_2H_3Cl_3O_2\times \text{Molar mass of }C_2H_3Cl_3O_2$

$\text{Mass of }C_2H_3Cl_3O_2=(2.0\times 10^{-2}mole)\times (165.5g/mole)=3.31g$

The mass in grams of $2.0\times 10^{-2}$ mole chloral hydrate is, 3.31 g

(d) To calculate the number of chlorine atoms are in 5.0 g chloral hydrate.

First we have to determine the moles of $C_2H_3Cl_3O_2$ .

$\text{Moles of }C_2H_3Cl_3O_2=\frac{\text{Mass of }C_2H_3Cl_3O_2}{\text{Molar mass of }C_2H_3Cl_3O_2}=\frac{5g}{165.5g/mole}=0.03moles$

Now we have to calculate the number of chlorine atoms in chloral hydrate.

In $C_2H_3Cl_3O_2$ , there are, 2 carbon atoms, 3 hydrogen atoms, 3 chlorine atoms and 2 oxygen atoms.

As, 1 mole of $C_2H_3Cl_3O_2$ contains $3\times 6.022\times 10^{23}$ chlorine atoms

So, 0.03 mole of $C_2H_3Cl_3O_2$ contains $0.03\times 3\times 6.022\times 10^{23}=5.4\times 10^{22}$ chlorine atoms

The number of chlorine atoms in 5.0 g chloral hydrate is, $5.4\times 10^{22}$

(e) To calculate the mass of chloral hydrate would contain 1.0 g Cl.

As, $3\times 35.5g$ of chlorine present in 165.5 g of $C_2H_3Cl_3O_2$

So, 1 g of chlorine present in $\frac{165.5}{3\times 35.5}=1.55g$ of $C_2H_3Cl_3O_2$

The mass of chloral hydrate will be, 1.55 g

(f) To calculate the mass of exactly 500 molecules of chloral hydrate.

As, $6.022\times 10^{23}$ molecules of chloral hydrate has 165.5 g mass of chloral hydrate

So, 500 molecules of chloral hydrate has $\frac{6.022\times 10^{23}}{500}\times 165.5=1.99\times 10^{23}$ mass of chloral hydrate

The mass of exactly 500 molecules of chloral hydrate is, $1.99\times 10^{23}$

3 0

3 years ago

The concentration of dye in Solution A is 25.527 M. You have 13 mL of water at your disposal to make the dilutions. The solution

Lady bird [3.3K]

Answer:

38825.472 M

Explanation:

Using

C1V1=C2V2

To make solution B,

C1 = 25.527 M, V1 = 13 mL

C2 = 25.527 x 8 = 204.216 M

V2 = 13 X 12 = 156 mL

To make solution C'

C1 = 8 X 202.216 M = 1617.728

V1 = 8 X 156mL = 1248 mL

V2 = 4 X 13 = 52 mL

C2 = C1V1 / V2

=1617.728 X 1248 / 52

=38825.472 M

7 0

4 years ago

A sample of gas is held at 1000C at a volume of 20 L. If the volume is increased to 40 L, what is the new temperature of the gas

kaheart [24]

Answer:

The new temperature will be 2546 K or 2273 °C

Explanation:

Step 1: Data given

The initial temperature = 1000 °C =1273 K

The volume = 20L

The volume increases to 40 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 20L

⇒with T1 = the initial temperature = 1273 K

⇒with V2 = the increased volume = 40L

⇒with T2 = the new temperature = TO BE DETERMINED

20L/ 1273 K = 40L / T2

T2 = 40L / (20L/1273K)

T2 = 2546 K

The new temperature will be 2546 K

This is 2546-273 = 2273 °C

Since the volume is doubled, the temperature is doubled as well

8 0

3 years ago

yanalaym [24]

Answer:

Explanation:

hope it helpls plz name brainlest

8 0

3 years ago

You add 21.00 ml of a 1.50 m solution to a 100 ml volumetric flask and then fill the flask with water to the 100 ml mark. what i

Furkat [3]

The new concentration of sodium phosphate is 0.315 mol·L⁻¹.

We can use the dilution formula :

c₁V₁ = c₂V₂

We can rearrange the formula to get

c₂= c₁ × (V₁/V₂)

∴ c₂ = 1.50 mol·L⁻¹ × (21.00 mL)/(100.0 mL)] = 1.50 mol·L⁻¹ × 0.2100

= 0.315 mol·L⁻¹

6 0

4 years ago