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Anika [276]
3 years ago
6

What mass of NaOH is in 250 mL of a 3.0M NaOH solution​

Chemistry
2 answers:
Kay [80]3 years ago
7 0

Answer:

about 30g

Explanation:

Molarity = mol/L

*convert mL to L*

3M=xmol/.25

xmol = .75

*convert to g

.75 mol NaOH * 40g/mol = 30 g

vodka [1.7K]3 years ago
6 0

Explanation:

Explanation: It's a 3.0 M solution so 1 litre of solution contains 3 moles of NaOH. 250 ml of solution therefore contains 0.25 x 3 = 0.75 moles of NaOH. The molar mass of NaOH is 39.9971 g/mol, so 250 ml of this solution contains 0.75 x 39.9971 = 29.99 g, or if you round it up 30.0 g.

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Which is also known as artificial (synthetic) silk​
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Why were colonists in new england angry at king james 2 abd governor edmund andros of new england?​
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3 years ago
Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
8 0
3 years ago
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