The ratio of buffer C₂H₃O₂ /HC₂H₃O₂ must you use are1:0.199 or 10:2
the ratio of buffer C₂H₃O₂ /HC₂H₃O₂ can be calculate using the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:
pH = pKa + log ([base]/[acid]
Where,
[base] = concentration of C₂H₃O₂in molarity or moles
[acid] = concentration of HC₂H₃O₂ in molarity or moles
For the sake of easy calculation, allow us to assume that:
[base] =1
[acid] = x
Therefore using equation 1,
5.44 = 4.74 + log (1 / x)
log [base / acid] = 0.7
1 / x = 5.0118
x = 0.199
The required ratio of buffer C₂H₃O₂ /HC₂H₃O₂ is 1:0.199 or 10:2
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Answer:
Explanation:
A solid, at a given temperature, has a definite volume
and shape which may be affected by changes in
temperature. Solids usually increase slightly in size
when heated (expansion) and usually
decrease in size if cooled (contraction).A liquid, at a given temperature, has a
fixed volume and will take up the shape of any
container into which it is poured. Like a solid, a
liquid’s volume is slightly affected by changes in
temperature.
A gas, at a given temperature, has neither a definite
shape nor a definite volume. It will take up the shape
of any container into which it is placed and will
spread out evenly within it. Unlike those of solids
and liquids, the volumes of gases are affected quite
markedly by changes in temperature.
Sodium represents by the (Na).
Considering the Boyle's law, the new pressure of the sample is 1,776 mmHg.
<h3>What is Boyle's law</h3>
Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.
Boyle's law states that the volume occupied by a given mass of gas at constant temperature is inversely proportional to the pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
P×V=k
Now it is possible to assume that you have a certain volume of gas V1 which is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and the following will be true:
P1×V1=P2×V2
<h3>New pressure</h3>
In this case, you know:
- P1= 740 mmHg
- V1= 3 L
- P2= ?
- V2= 1.25 L
Replacing in Boyle's law:
740 mmHg× 3 L=P2× 1.25 L
Solving:
P2= (740 mmHg× 3 L) ÷ 1.25 L
P2= 1,776 mmHg
Finally, the new pressure of the sample is 1,776 mmHg.
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Answer:
≈170.22 [gr].
Explanation:
1) M(Cu₂CO₃)=2*64+12+3*16=188 [gr/mol];
2) the part of copper in Cu₂CO₃ is:
3) 16/47% of 500 [gr] is:
![500*\frac{16}{47}=170.22[gr].](https://tex.z-dn.net/?f=500%2A%5Cfrac%7B16%7D%7B47%7D%3D170.22%5Bgr%5D.)
