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3 years ago

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A chemist prepares a solution by adding 448 mg of Co(NO3)2 (MW = 182.94 g/mol ) to a volumetric flask, and then adding water unt

il the total volume of the contents of the flask reaches the calibration line that indicates 500 mL . Determine the molarity of the prepared solution.

1 answer:

Ghella [55]3 years ago

6 0

Answer:

Molarity of solution is 0.0049 M

Explanation:

Molarity of a solution is the ratio of number of moles of solute to volume of solution in Liter

Number of moles is the ratio of mass to molecular weight of a compound.

Here $Co(NO_{3})_{2}$ is the solute.

448 mg of $Co(NO_{3})_{2}$ = 0.448 g of $Co(NO_{3})_{2}$

So, 0.448 g of $Co(NO_{3})_{2}$ = $\frac{0.448}{182.94}$ moles of $Co(NO_{3})_{2}$ = 0.00245 moles of $Co(NO_{3})_{2}$

Total volume of solution = 500 mL = 0.5 L

So, molarity of solution = $\frac{0.00245}{0.5}M=0.0049 M$

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A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine

bagirrra123 [75]

Answer:

The final volume of the sample of gas $V_{2}$ = 0.000151 $m^{3}$

Explanation:

Initial volume $V_{1}$ = 200 ml = 0.0002 $m^{3}$

Initial temperature $T_{1}$ = 296 K

Initial pressure $P_{1}$ = 101.3 K pa

Final temperature $T_{2}$ = 336 K

Final pressure $P_{2}$ = K pa

Relation between P , V & T is given by

$P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }$

Put all the values in the above equation we get

$101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )$

$V_{2}$ = 0.000151 $m^{3}$

This is the final volume of the sample of gas.

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4 years ago

What fraction of a Sr-90 sample remains unchanged after 87.3 years

jolli1 [7]

The answer is 1/8.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. $(1/2)^{n} = x$ ,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. $t_{1/2} = \frac{t}{n}$
where:
<span> $t_{1/2}$ - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
<span> $t_{1/2}$ = 28.8 years

We need:
n = ?
x = ?
</span>
We could first use the second equation, to calculate n:
<span>If:
$t_{1/2} = \frac{t}{n}$ ,
</span>Then:
$n = \frac{t}{ t_{1/2} }$
⇒ $n = \frac{87.3 years}{28.8 years}$
⇒ $n=3.03$
<span>⇒ n ≈ 3
</span>
Now we can use the first equation to calculate the remained amount of the sample.
<span> $(1/2)^{n} = x$
</span>⇒ $x=(1/2)^3$
⇒ $x= \frac{1}{8}$ <span>
</span>

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3 years ago

Read 2 more answers

In the presence of CN-, Fe3 forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3 and Fe(CN)63- are 8.5 x 10-40

aleksklad [387]

Answer:

Kf = 9.96x10⁴¹

Explanation:

When Fe³⁺ and CN⁻ are in water, complex Fe(CN)₆³⁻ is formed, thus:

Fe³⁺ + 6CN⁻ ⇄ Fe(CN)₆³⁻

Kf is defined as:

Kf = [Fe(CN)₆³⁻] / [Fe³⁺] [CN⁻]⁶

Equilibrium concentration of ions is:

[Fe(CN)₆³⁻] = 1.5x10⁻³M

[Fe³⁺] = 8.5x10⁻⁴⁰

[CN⁻] = [KCN] = 0.11M

Replacing in Kf expression:

Kf = [1.5x10⁻³M] / [8.5x10⁻⁴⁰] [ 0.11M]⁶

<h3>Kf = 9.96x10⁴¹</h3>

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