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3 years ago

How many liters of a 4.0 M CaCl2 solution would contain 2 moles of CaCl2?

Chemistry

1 answer:

nekit [7.7K]3 years ago

5 0

Answer:

The answer to your question is 0.5 liters

Explanation:

Data

[CaCl₂] = 4.0 M

number of moles = 2

volume = ?

Process

To solve this problem use the formula of Molarity and solve it for volume, substitute the values and simplify.

-Formula

Molarity = moles / volume

-Solve for volume

Volume = moles / molarity

-Substitution

Volume = 2/4

-Simplification

Volume = 0.5 liters.

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4 years ago

If molten sodium chloride is split by electrolysis, which reaction occurs at the anode?

mariarad [96]

Answer: $2Cl^-(l)\rightarrow 2e^-+Cl_2(g)$

Explanation:

Electrolysis of a subastance is breaking it into its constituents by the action of electrical current.

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$2NaCl(l)\rightarrow 2Na^++2Cl^-$

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3 years ago

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3 years ago

What causes the output work of a machine to be less than the input work?????

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Inefficiency of the material or engine causes the output work of a machine to be less than the input work. It is governed by the second law of thermodynamics wherein no product of work can produce 100 percent efficiency. Only the Carnot engine can produce a 100 percent effieciency.

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4 years ago

A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region o

Llana [10]

Answer:

5.75

Explanation:

Weak acid and strong base,

Make it a two part problem

The first will be Partial neutralization of the weak acid, while the second is the equilibrium and final pH

1.Parameters

HC2H3O2 =10mL x 0.75M = 7.5 mmol

NaOH =22mL x 0.30M = 6.6 mmol

R HC2H3O2 + OH- --> C2H3O2- + H2O

I 7.5 mmol 6.6 mmol 0 mmol ignore

C -6.6 mmol -6.6 mmol +6.6 mmol

E 0.9 mmol 0.0 mmol 6.6 mmol

2.) HC2H3O2 --> H+ + C2H3O2-

Initial concentrations

[HC2H3O2]: 0.9 mmol / 32mL = 0.0281M

[H+]:

[C2H3O2-]: 6.6 mmol x 32mL = 0.206M

Concentrations at equilibrium

[HC2H3O2]: 0.0281M - x

[H+]: [C2H3O2-]: 0.206M + x

If x is small and can be ignored except [H+]

Substitute and solve

1.3 x 10-5 = (x)(0.206)

(0.0281)

X= 1.77 x 10-6M => pH = 5.75

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3 years ago