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4 years ago

If 10.g of AgNo3 is available, what volume of 0.25 M AgNo3 can be prepared

1 answer:

3 0

The equation is L = m/M
First, covert 10. grams of AgNO3 to moles which is 0.059 moles.
Divide 0.059 moles by 0.25M which is 0.24 liters.

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You pour 25.0 mL of 6.70 M NaOH stock solution into your beaker. How much water will you have if it is diluted to 3.40 M

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3 years ago

Oxalic acid is a polyprotic acid. Write balanced chemical equations for the sequence of reactions that oxalic acid can undergo w

tiny-mole [99]

The chemical formula of oxalic is $H_{2}C_{2}O_{4}$

When oxalic acid reacts with water, first, oxalic acid removes one proton and results in the formation of mono acids.

After that, in second step, oxalic acid in aqueous solution removes another proton which shows it is a polyprotic acid.

The chemical equations are: (the reactions occurs in two steps due to presence to hydrogen atoms).

When one proton is removed:

$H_{2}C_{2}O_{4} (aq)+H_{2}O (l)\rightarrow HC_{2}O_{4}^{-}(aq)+H_{3}O^{+}(aq)$

When another proton is removed:

$HC_{2}O_{4}^{-} (aq)+H_{2}O (l)\rightarrow C_{2}O_{4}^{2-}(aq)+H_{3}O^{+}(aq)$

The dissociation of oxalic acid in water in shown in the image.

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3 years ago

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maria [59]

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3 years ago

Perform the following operation

k0ka [10]

Answer:

1449 × 10 ^-5

Explanation:

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3 0

3 years ago

what is the pH of a solution that results when 0.010mol HNO3 is added to 500.ml of a solution that is 0.10M in aqueous ammonia a

WARRIOR [948]

Answer : The

pH of a solution is, 8.56

Explanation : Given,

$K_b=1.8\times 10^{-5}$

Concentration of ammonia (base) = 0.10 M

Concentration of ammonium nitrate (salt) = 0.55 M

First we have to calculate the value of $pK_b$ .

The expression used for the calculation of $pK_b$ is,

$pK_b=-\log (K_b)$

Now put the value of $K_b$ in this expression, we get:

$pK_b=-\log (1.8\times 10^{-5})$

$pK_b=5-\log (1.8)$

$pK_b=4.7$

Now we have to calculate the pOH of buffer.

Using Henderson Hesselbach equation :

$pOH=pK_b+\log \frac{[Salt]}{[Base]}$

Now put all the given values in this expression, we get:

$pOH=4.7+\log (\frac{0.55}{0.10})$

$pOH=5.44$

The pOH of buffer is 5.44

Now we have to calculate the pH of a solution.

$pH+pOH=14\\\\pH+5.44=14\\\\pH=14-5.44\\\\pH=8.56$

Thus, the pH of a solution is, 8.56

8 0

3 years ago