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kotegsom [21]
3 years ago
14

What is the percent by mass of nahco3 in a solution containing 10 g of nahco3 dissolved in 400 ml of h2o?

Chemistry
1 answer:
Marat540 [252]3 years ago
7 0
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
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guajiro [1.7K]
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8 0
3 years ago
Read 2 more answers
A compound is composed of 58.8% c, 9.8% h, and 31.4% o, and the molar mass is 102 g/mol. what is the molecular formula for this
Ivahew [28]

The solution would be like this for this specific problem:

 

<span>Moles of carbon = 58.8 / 12 = 4.9 </span><span>
<span>Moles of hydrogen = 9.8 / 1 = 9.8 </span>
<span>Moles of oxugen = 31.4 / 16 m= 1.96 </span>
<span>Ratio 4.9 / 1.96 = 2.5 9.8 / 1.96 = 5.0 1.96 / 1.96 = 1 </span></span>

Simplest formula = C5H10<span>


</span><span>I hope this helps and if you have any further questions, please don’t hesitate to ask again.</span>
8 0
3 years ago
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What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?
Ymorist [56]

Answer:

\% diss = 50\%

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}

Thus, it is possible to find x given the pH as shown below:

x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M

So that we can calculate the initial concentration of the acid:

\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\

[HA]_0=3.64x10^{-5}M

Therefore, the percent dissociation turns out to be:

\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%

Best regards!

6 0
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pentagon [3]

Answer:

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Explanation:

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6 0
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