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bija089 [108]
3 years ago
12

A chemist makes 620 mL of barium chloride working solution by adding distilled water to180mL of a 1.39mol/L stock solution of ba

rium chloride in water. Calculate the concentration of the chemist's working solution. Be sure your answer has the correct number of significant digits.
Chemistry
2 answers:
katen-ka-za [31]3 years ago
7 0

Answer:

The correct answer is 0.40 mol/L.

Explanation:

In order to solve the problem, we use the following expression, where V1 and C1 are the volume and concentration of stock solution and V2 and C2 are the volume and concentration of working solution:

V1 x C1= V2 x C2

⇒C2= V1 x C1/ v2

  C2= (180 ml x 1.39 mol/L)/ 620 ml

  C2= 0.40 mol/L

Svetllana [295]3 years ago
6 0

Answer:

0.404 mol / L

Explanation:

The <em>total moles of barium chloride</em> can be calculated from <em>the volume and concentration of the stock solution</em>:

1.39 mol/L * 0.180 L = 0.2502 moles barium chloride

Now we <u>calculate the concentration of the working solution, by dividing the total moles by the final volume</u>:

0.2502 mol / 0.620 L = 0.404 mol / L

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Mass of H_{2} that would produce 208 kg methanol =

6492 mol CH_{3}OH * \frac{2 mol H_{2}}{1 mol CH_{3}OH} * \frac{2.02 g H_{2}}{1 mol H_{2}} * \frac{1 kg}{1000 g}=  26.2 kg H_{2}

2) Mass of sodium = 1.3 * 10^{21} atoms Na * \frac{1 mol Na}{6.022* 10^{23}atoms Na} * \frac{22.99 g Na}{1 mol Na} = 0.0496 g Na

3) Mass of mercury = 1.2 mL * 13.6 \frac{g}{mL} = 16.32 g

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3 years ago
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yanalaym [24]
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3 years ago
How many atoms are there in 2.43 g of calcium?
Evgen [1.6K]

Answer:

Explanation:

We're asked to calculate the number of atoms of

Ca

in

153

g Ca

.

What we must first do is convert the given mass of calcium to moles of calcium, using its molar mass (referring to a periodic table, this is

40.08

g

mol

):

153

g Ca

(

1

mol Ca

40.08

g Ca

)

=

3.82

mol Ca

Using Avogadro's number,

6.022

×

10

23

particles

mol

, we can calculate the number of atoms present:

3.82

mol Ca

(

6.022

×

10

23

atoms Ca

1

mol Ca

)

=

2.30

×

10

24

atoms Ca

3 0
2 years ago
Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate
nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

N=N_A\times n

Charge on N electrons : Q

Q = N\times 1.602\times 10^{-19} C

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A

18.0 Ampere is the size of electric current that must flow.

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Leokris [45]
2 5 3 4 1 is the answers
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3 years ago
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