At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.

Answers

Answer 1

Answer:

Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_(H_2S)=K_H* p_(liquid)$

where,

$K_H$ = Henry's constant = $0.087M/atm$

$p_(H_2S)$ = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

$C_(H_2S)=0.087M/atm* 2.42atm\n\nC_(H_2S)=0.2105M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.2105 M

Molar mass of hydrogen sulfide = 34 g/mol

Volume of solution = 400.0 mL

Putting values in above equation, we get:

$0.2105M=\frac{\text{Mass of hydrogen sulfide}* 1000}{34g/mol* 400.0mL}\n\n\text{Mass of }H_2S=(0.2105* 34* 400)/(1000)=2.86g$

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Related Questions

What is the value for the kinetic energyfor a n = 2 Bohr orbit electron in Joules?
If the heat of combustion of carbon monoxide (CO) is −283.0 kJmole, how many grams of carbon monoxide must combust in order to release 2,500.kJ of energy?
The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 7.300 g C 6 H 6 7.300 g C6H6 is burned and the heat produced from the burning is added to 5691 g 5691 g of water at 21 ∘ 21 ∘ C, what is the final temperature of the water?
Predict the products and write balanced net ionic equations for the following reactions. (g) SnCl2 is added to KMnO4 solution (acidic) forming Mn2
2.67 grams of butane (C4H10) is combusted in a bomb calorimeter. The temperature increases from 25.68 C to 36.2C. What is the change in the internal energy (deltaE) in KJ/mol for the reaction if the heat capacity of the bomb calorimeter is 5.73 kJ/C?

Isn't this false? For the industrial production of indigo carmine, a blue food colouring additive, a synthetic process with an E-factor of 17.4 produces less waste than a synthetic process with an E-factor of 3.0.

The answer I got was False, is this correct?

Answers

Answer: yes it is false

Explanation:

The statement is false. A synthetic process with a lower E-factor produces less waste than a process with a higher E-factor.

The E-factor is a measure of the waste generated during a manufacturing process. It is calculated by dividing the total mass of waste produced by the mass of the desired product. A lower E-factor indicates that less waste is generated per unit of product.

In this case, the synthetic process with an E-factor of 3.0 produces less waste than the process with an E-factor of 17.4. This means that the process with an E-factor of 3.0 is more efficient in terms of waste reduction.

To create an image, the lens of an eye ________focuses light on the cornea
reflects light away from the retina
bends the cornea to correct vision
focuses light on the retina

Answers

It focuses light on the retina.

Answer:

focuses light on the retina

Explanation:

What is the toto number of neutrons in an atom of 73 li

Answers

The 73 in Li means that there are 73 protons and neutrons in this atom. To find the number of neutrons, simply subtract the number of protons, which is equal to the atomic number (which is 3 in this case).

70 neutrons.

An alloy is a mixture of metals to obtain a more durable and strongersubstance.
True
False

Answers

Answer: False

Explanation: Alloys are harder and stronger because the different-sized atoms of the mixed metals make the atomic layers less regular, so they cannot slide as easily.

The amount of I − 3 ( aq ) in a solution can be determined by titration with a solution containing a known concentration of S 2 O 2 − 3 ( aq ) (thiosulfate ion). The determination is based on the net ionic equation 2 S 2 O 2 − 3 ( aq ) + I − 3 ( aq ) ⟶ S 4 O 2 − 6 ( aq ) + 3 I − ( aq ) Given that it requires 29.4 mL of 0.380 M Na 2 S 2 O 3 ( aq ) to titrate a 30.0 mL sample of I − 3 ( aq ) , calculate the molarity of I − 3 ( aq ) in the solution.

Answers

Answer:

The molarity of I₃⁻ (aq) solution: M₂ = 0.186 M

Explanation:

Given net ionic equation:

2S₂O₃²⁻ (aq) + I₃⁻ ( aq ) ⟶ S₄O₆²⁻ (aq) + 3I⁻ (aq)

Number of moles of S₂O₃²⁻: n₁ = 2, Number of moles of I₃⁻: n₂ = 1

Given- For S₂O₃²⁻ solution: Molarity: M₁ = 0.380 M, Volume: V₁ = 29.4 mL;

For I₃⁻ (aq) solution: Molarity: M₂ = ? M, Volume: V₂ = 30.0 mL

To calculate the molarity of I₃⁻ (aq) solution, we use the equation:

$(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))$

$((0.380 M)* (29.4 mL))/(2)=(M_(2)* (30.0 mL))/(1)$

$\Rightarrow M_(2) = ((0.380 M)* (29.4 mL))/((30.0 mL)* 2) = 0.186 M$

Therefore, the molarity of I₃⁻ (aq) solution: M₂ = 0.186 M

How many iron atoms are in 0.32 mol of Fe2031? 3.9x 1023 jron atoms O 3.9 iron atoms O 6.02 x 1023 iron atoms 1.9 x 1023 iron atoms O 11x 10-24 iron atoms

Answers

Answer: $3.9* 10^(23)$ iron atoms

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023* 10^(23)$ of particles.

$1 molecule of [tex]Fe_2O_3$ contains= 2 atoms of iron

$1 mole of [tex]Fe_2O_3$ contains= $2* 6.023* 10^(23)=12.05* 10^(23)$ atoms of iron

thus 0.32 moles of $Fe_2O_3$ contains= $(12.05* 10^(23))/(1)* 0.32=3.9* 10^(23)$ atoms of iron

Thus the sample would have $3.9* 10^(23)$ iron atoms.

Other Questions

In what way(s) is(are) the structure of water similar to that of CHCl3? In what way is the structure of water different from that of CHCl3?

What is the only gaseous component of the atmosphere whose concentration is routinely contained in all weather reports?

How many grams of solid sodium cyanide should be added to 1.00 L of a 0.119 M hydrocyanic acid solution to prepare a buffer with a pH of 8.809

Perform the calculation and report the answer using the proper number of significant figures. Make sure the answer is rounded correctly. 1.012×10^-3 J/(0.015456 g)(298.3682−298.3567)K