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SSSSS [86.1K]
3 years ago
8

DUPLICATE. The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to

form its conjugate, and the other half still remains unreacted. If 0.580 moles of a monoprotic weak acid (Ka = 4.5 _ 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?
Chemistry
1 answer:
Law Incorporation [45]3 years ago
5 0
At the half equivalence point [HA] = [A-] and pH = pKa 

<span>if Ka is 5.2e-5 then pKa = pH = 4.28</span>
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Which of the following is an example of a neutralization reaction?
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HCl (aq) + NaOH (aq)→H2O (l) + NaCl (aq)
Acid reacts with base to give H2O and salt
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3 years ago
Science<br><br> 0.00098 I need that converted to the scientific notation I need help quick
Viktor [21]

Answer:

9.8×10^-4...... is the answer

7 0
2 years ago
What is the mass of 2.9 moles of calcium? Explain please
Alik [6]

Answer:

116 g

Explanation:

From the question given above, the following data were obtained:

Number of mole of calcium = 2.9 moles

Mass of calcium =.?

The mole and mass of a substance are related according to the following formula:

Mole = mass / molar mass

With the above formula, we can obtain the mass of calcium. This can be obtained as follow:

Number of mole of calcium = 2.9 moles

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4 0
3 years ago
A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr
umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

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[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

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This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0
3 years ago
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Answer:

B: The sulfuric acid is not consumed or react with the reactant.

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