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kompoz [17]
3 years ago
13

Plz help!!! will mark as brainliest!!!!!

Chemistry
1 answer:
Ivenika [448]3 years ago
4 0
The answer is B. Western Line
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An aqueous solution has a molality of 3. 64 kcl. what is the the mass percent of kcl in the solution? enter your answer to three
Sliva [168]

The mass percentage of KCl solution is 21.34%

Calculation,

Given molality of KCl aqueous solution =  3. 64  m =  3. 64 mole/ Kg

It means 3. 64 mole of solute present in 1Kg of solvent

or ,  3. 64 mole of solute present in 1000 g of solvent ( water )

Mass of solute ( KCl ) =  3. 64 mole×74.55 g/mole =271.3 g

The total mass of solution ( KCl + water ) = 271.3 g + 1000 g = 1271.3 g

Mass percentage is equal to percentage of mass of solute present in total mass of solution.

Mass percentage = 100×271.3 g/1271.3 g = 21.34%

to learn more about mass percentage

brainly.com/question/27108588

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5 0
2 years ago
(a) Consider the reaction of hydrogen sulfide with methane, given below:
Burka [1]

Answer:

a. H2S(g)/t = 1.48 mol/s

CS2(g)/t = 0.740mol/s

H2(g)/t = 2.96mol/s

b. Ptot /t = 981torr/min

Explanation:

a. Based on the reaction:

CH4(g) + 2 H2S(g) → CS2(g) + 4 H2(g)

<em>1 mole of CH4 reacts with 2 moles of H2S producing 1 mole of CS2 and 4 moles of 4H2</em>

<em />

If CH4 decreases at the rate of 0.740mol/s, H2S decreases twice faster, that is 0.740mol/s = 1.48 mol/s

CS2 is produced with the same rate of CH4 because 1 mole of CH4 produce 1 mole of CS2 = 0.740mol/s

The H2 is produced four times faster than CH4 is decreased, that is:

0.740mol/s * 4 = 2.96mol/s

b. With the reaction:

2 NH3(g) → N2(g) + 3 H2(g)

2 moles of ammonia are consumed whereas 1 mole of N2 and 3 moles of H2 are produced.

That means 2 moles of gas are consumed and 4 moles of gas are produced.

If the NH3 decreases at a rate of 327torr/min, the gases are produced in a rate twice faster. That is 327torr/min*2 =

654torr/min

The rate of change of the total pressure is rate of reactants + rate of products:

654torr/min + 327torr/min =

981torr/min

6 0
4 years ago
In the double replacement reaction between na2so4 (aq) and bacl2 (aq), will a precipitate form?
bagirrra123 [75]
A precipitate will form since BaSO₄ is insoluble in water.
8 0
4 years ago
27.8 mL solution of 0.797 M HCHO2 with 0.928 M NaOH. What is the pH for the solution at the equivalence point in the titration?
kati45 [8]

Answer:

8.69 is the pH at the equivalence point

Explanation:

Formic acid, HCHO₂, reacts with NaOH as follows:

HCHO₂ + NaOH → NaCHO₂ + H₂O.

At the equivalence point you will have in the reaction just NaCHO₂ and H₂O. The concentration of NaCHO₂ will be:

<em />

<em>Moles: </em>0.0278L * 0.797mol/L = 0.02216moles

To reach the equivalence point it is necessary to add:

0.02216mol * (1L / 0.928mol) = 0.0239L

Total volume in the equivalence point:

0.0278L + 0.0239L = 0.0517L

Concentration: 0.02216moles / 0.0517L = 0.429M

The equilibrium of NaCHO₂, CHO₂⁻, in water is:

CHO₂⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HCHO₂(aq)

Where Kb, 5.56x10⁻¹¹ is defined as:

5.56x10⁻¹¹ = [OH⁻] [HCHO₂] / [CHO₂⁻]

In the equilibrium, it is produced X OH⁻ and HCHO₂, and as concentration of NaCHO₂ is 0.429M:

5.56x10⁻¹¹ = [X] [X] / [0.429M]

2.383x10⁻¹¹ = X²

4.88x10⁻⁶ = X = [OH⁻]

As pOH = -log [OH⁻]

pOH = 5.31

And pH = 14 - pH

pH = 8.69 is the pH at the equivalence point

3 0
4 years ago
Which type of change occurs when electrons form a bond between two atoms?
Nataly [62]
B is the answer. hope this helps
8 0
4 years ago
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