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denis-greek [22]

3 years ago

8

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M

. At equilibrium, [Cl2]= 1.3×10−2M .

1 answer:

lukranit [14]3 years ago

5 0

I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.

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musickatia [10]

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What is the rate of a reaction if the value of kis 0.1, [A] is 1 M, and [B] is 2 M?

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What is the molarity when 8.0 moles of solute are dissolved in 1.5 liters of solution?

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3 years ago

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Dennis_Churaev [7]

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5 0

3 years ago

1) Sodium hydroxide is deliquescent. A sample of 3.0 g was dissolved in 100 mL; 10 mL was titrated with 34.9 mL HCl 0.2 M. What

ELEN [110]

From the calculation, the percentage of water in the sodium hydroxide sample is 7%.

<h3>What is neutralization?</h3>

The term neutralization has to do with the reaction between an acid and a base to yiled salt and water.

Now we have to apply the titration formula;

CAVA/CBVB = NA/NB

CA = concentration of acid

CB = concentration of base

VA = volume of acid

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

The reaction equation is; HCl + NaOH ----->NaCl + H2O

CAVANB = CBVBNA

CB = CAVANB /VBNA

CB = 34.9 * 0.2 M * 1/10 * 1

CB = 0.698 M

Number of moles = Conncentration * volume

= 0.698 M * 100/1000 L = 0.0698 moles

Mass = Number of moles * molar mass

Mass = 0.0698 moles * 40 g/mol = 2.79 g

percent of NaOH = 2.79 g/ 3g * 100/1 = 93%

Percent of water = 100- 93 = 7%

Learn more about neutralization: brainly.com/question/15395418

5 0

2 years ago