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nignag [31]
2 years ago
11

When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plo

t of the titration will have Group of answer choices

Chemistry
1 answer:
dezoksy [38]2 years ago
4 0

Complete question is;

When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have

a. a pH of 7 at the equivalence point.

b. two equivalence points below 7.

c. no equivalence point.

d. one equivalence point.

e. two distinct equivalence points

Answer:

Option E - Two Distinct Equivalence points

Explanation:

I've attached a sample diprotic acid titration curve.

In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.

Thus, it is clear that there are two distinct equivalence points.

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How many moles are equal to 5.82 x 10^23 atoms of tungsten (w)
fiasKO [112]

Answer:

0.97 mole

Explanation:

1 mole will give 6.02×10^23 atoms

Xmole of tungsten will give 5.82×10^23 atom of tungsten

X= 5.82×10^23/ 6.02×10^23

X = 0.97 moles of tungsten

6 0
2 years ago
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Many plants are poisonous because their stems and leaves contain oxalic acid, H2C2O4, or sodium oxalate, Na2C2O4. When ingested,
Neporo4naja [7]

Answer:

Explanation:

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CaCl₂(aq) ----> Ca²⁺(aq) + 2Cl⁻(aq)

Similarly, sodium oxalate when dissolved in water dissociates into sodium and oxalate ions.

Na₂CO₄(aq) ----> 2Na⁺(aq) + C₂O₄²⁻(aq)

However, in a double displacement reaction where the two solutions of the salts are mixed, the insoluble salt calcium oxalate is precipitated. The net ionic equation for the reaction is shown below:

Ca²⁺(aq) + C₂O₄²⁻(aq) ----> CaC₂O₄(s)

8 0
2 years ago
What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL
tatiyna

Answer:

The answer is "\bold{\log \frac{[0] mole}{[R]mole}}"

Explanation:

E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\

In the above-given equation, we can see from E_{ceu}, of both oxidant conc^nas well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor  and each other suspend

\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\

\to {\log \frac{[0] mole}{[R]mole}}

3 0
2 years ago
If 2.38 mol of a gas has a volume of 120.0 mL, what is the volume of 1.97 mol of the gas at the same temperature and pressure?
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According to the Avogadro's law, equal volumes of all gases contain equal number of moles under the same conditions of temperature and pressure.

Here the initial volume, V_{1} = 120.0 mL

Initial moles of the gas, n_{1}=2.38 mol

Final volume of the gas V_{2} = ?

Final moles of the gas n_{2}= 1.97 mol

Plugging in the Avogadro's equation,

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}

\frac{120.0 mL}{2.38 mol} = \frac{V_{2}}{1.97 mol}

V_{2} = 99.33 mL

Therefore the 1.97 mol of the gas occupies 99.33 mL at constant temperature and pressure.

7 0
2 years ago
Jack researched J.J. Thomson’s experiments. Read Jack’s summary below and find the error.
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Answer:

D. Error is in sentence 1. The beam was a cathode ray.

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