All categories

3 years ago

At 9°C a gas has a volume of 6.17 L. What is its volume when the gas is at standard temperature?

Chemistry

1 answer:

Alex17521 [72]3 years ago

5 0

Answer:

V₂ = 5.97 L

Explanation:

Given data:

Initial temperature = 9°C (9+273 = 282 K)

Initial volume of gas = 6.17 L

Final volume of gas = ?

Final temperature = standard = 273 K

Solution:

Formula:

The Charles Law will be apply to solve the given problem.

According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 6.17 L × 273K / 282 k

V₂ = 1684.41 L.K / 282 K

V₂ = 5.97 L

You might be interested in

Pls help the answers and questions r in the pic

soldier1979 [14.2K]

It would be 9*10 to the -3

6 0

3 years ago

Read 2 more answers

If you need to measure the mass of a greeting card you are sending to a friend through the mail, what unit of measurement would

erica [24]

You would use grams to measure ot

4 0

3 years ago

Which of the following is a prokaryote that contains a cell wall, cell membrane, and flagellum?

Reil [10]

Answer:

Bacteria

Explanation:

hope this helps please gimme brainliest

6 0

3 years ago

Valeric acid, HC5H9O2 (Ka = 1.5 ✕ 10−5), is used in the manufacture of magnesium valerate, a nerve-calming agent. What is the hy

Studentka2010 [4]

Answer:

$[H^+]=0.00332M$

Explanation:

Hello,

In this case, considering the dissociation of valeric acid as:

$HC_5H_9O_2 \rightleftharpoons C_5H_9O_2 ^-+H^+$

Its corresponding law of mass action is:

$Ka=\frac{[H^+][C_5H_9O_2^-]}{[HC_5H_9O_2]}$

Now, by means of the change $x$ due to dissociation, it becomes:

$Ka=\frac{(x)(x)}{0.737-x}=1.5x10^{-5}$

Solving for $x$ we obtain:

$x=0.00332M$

Thus, since the concentration of hydronium equals $x$ , the answer is:

$[H^+]=x=0.00332M$

Best regards.

7 0

3 years ago

Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra

maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg(OH)2 can be given as:

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²

Step 2: Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

X = 2.412 * 10^-4 mol/L = solubility in water

Step 3: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+ = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X *(0.130)²

5.61*10^-11 = X * (0.130)^2

X = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0

4 years ago