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2 years ago

What is the final concentration if water is added to 0.50 L of a 12 M NaOH solution to make 3.0 L of a diluted NaOH solution?

Chemistry

2 answers:

Kisachek [45]2 years ago

6 0

Answer:

2 M NaOH solution

Explanation:

Initial concentration of NaOH = 12 M

Initial volume of solution = 0.5

Initial number of moles = Molarity×Volume = 12×0.5 = 6 moles

Note: On dilution the number of moles does not change.

Initial number of moles = final number of moles.

Final number of moles = 6 moles

Final volume of solution = 3 L

Final concentration (in Molarity) = $\frac{Final\ no\ of\ moles}{final\ volume\ in\ L}=\frac{6}{3}= 2\ M$

SVETLANKA909090 [29]2 years ago

3 0

Answer:

The answer to your question is the second option: 2M NaOH solution

Explanation:

Data

Volume 1 = 0.5 l

Concentration 1 = 12 M NaOH

Volume 2 = 3 l

Concentration 2 = ?

Formula

V₁C₁ = V₂C₂

Solve for C₂

C₂ = V₁C₁ / V₂

Substitution

C₂ = (0.5)(12) / 3

Simplification and result

C₂ = 6 / 3

C₂ = 2 M

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How many particles are in 12.47 grams of NaCl?

jeka94

If we are talking about moles then the answer to that is 0.22

5 0

3 years ago

A 0.105 L sample of an unknown HNO 3 solution required 35.7 mL of 0.250 M Ba ( OH ) 2 for complete neutralization. What is the c

oksano4ka [1.4K]

Answer: 0.17M

Explanation:

The equation for the reaction is :

2HNO3 + Ba(OH)2 —> Ba(NO3)2 + 2H2O

From the balanced equation, we obtain :

nA = mole of acid = 2

nB = mole of the base = 1

From the question, we obtain:

Va = Vol. Of acid = 0.105L

Ma = conc. Of acid =?

Vb = Vol of base = 35.7 mL = 0.0357L

Mb = conc. of base = 0.25M.

We solve for the conc. of the acid using:

MaVa / Mb Vb = nA / nB

(Ma x 0.105) / (0.25x0.0357) = 2

Cross multiply to express in linear form. We have:

Ma x 0.105 = 0.25 x 0.0357 x 2

Divide both side by 0.105. We have

Ma = (0.25 x 0.0357 x 2) / 0.105

Ma = 0.17M

Therefore the concentration of the acid(HNO3) is 0.17M

3 0

3 years ago

Need help setting the problem up

Sveta_85 [38]

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

n = 4.0 moles

Therefore, the number of mole of the gas is 4.0 moles

5 0

3 years ago

A pure copper penny contains approximately 2.9×1022 copper atoms. Use the following definitions to determine how many ______ of

hammer [34]

Complete question is;

A pure copper penny contains approximately 2.9 × 10^(22) copper atoms.

1 doz = 12

1 gross = 144

1 ream = 500

1 mol = 6.022 × 10^(23)

Use these definitions to determine the following:

A) How many dozens of copper atoms are in a penny.

B) How many gross of copper atoms are in a penny

C) How many reams of copper atoms are in a penny.

D) how many moles of copper atoms are in a penny?

All answers can be rounded to two significant figures

Answer:

A) 2.4 × 10^(21) dozens

B) 2.01 × 10^(20) gross

C) 5.8 × 10^(19) reams

D) 0.048 mol

Explanation:

A) A dozen contains 12.

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/12 dozens = 2.42 × 10^(21).

In 2 significant figures, we have;

2.4 × 10^(21) dozens

B) 1 gross = 144

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/144 gross ≈ 2.01 × 10^(20) gross

C) 1 ream = 500

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/500 reams = 5.8 × 10^(19) reams

D) 1 mol = 6.022 × 10^(23)

Therefore, 2.9 × 10^(22) copper atoms will contain;

(2.9 × 10^(22))/(6.022 × 10^(23)) = 0.048 mol

6 0

3 years ago

A mixture with medium sized particles that do not settle out is called

Alexandra [31]

Answer: colloid

Explanation:

Colloids are solutions in which small sized particles are suspended throughout the solution as they do not settle own their own. Colloids are defined as the mixtures where the size of the particle is within the range of 2nm to 1000 nm. In these mixtures, physical boundary is seen between the dispersed phase and dispersed medium.

Colloids are solutions with particle size intermediate between true solutions and suspensions. Suspensions have large sized particles which settle when left undisturbed for sometime and thus can be filtered off easily. The particle size in colloids is less and hence they do not settle under the effect of gravity.

7 0

3 years ago