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3 years ago

If this container of liquid water was frozen into ice and then, thawed back into liquid what would the volume be?

Chemistry

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

Answer:

Explanation:

because it's is still the same amount even if you freeze it

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An acid with a p K a of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated f

olga2289 [7]

Explanation:

According to the Henderson-Hasselbalch equation,

pH = $pK_{a}$ + $\frac{log[A^{-}]}{[HA]}$

Given values are pH = 6, $pK_{a}$ = 8

Putting given values into the above equation as follows.

6 = 8 + $\frac{log [A^{-}]}{[HA]}$

$\frac{log[A^{-}]}{[HA]}$ = -2

$\frac{[A^{-}]}{[HA]}$ = antilog -2

= 0.01

But according to the question, we need protonated to deprotonated ratio of $\frac{[HA]}{[A^{-}]}$

$\frac{[HA]}{[A^{-}]}$ = $\frac{1}{0.01}$

$\frac{[HA]}{[A^{-}]}$ = 100

Thus, we can conclude that ratio of the protonated to the deprotonated form of the acid is $\frac{100}{1}$ .

7 0

3 years ago

What weight of sodium formate must be added to 4.00L of 1.00 M formic acid to produce a buffer solution that has a pH of 3.50?

tester [92]

Answer:

156,4 g of sodium formate

Explanation:

The pka of the formic acid is 3,74. Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA] (1)

Where A⁻ is the conjugate base (Formate) and HA is the formic acid.

4.00L of 1.00M formic acid contain:

4.00L × (1.00mol /L) = 4.00 moles

Replacing these moles, the desired pH and the pka value in (1):

3,50 = 3,74 log₁₀ [A⁻] / 4,00 moles

-0,24 = log₁₀ [A⁻] / 4,00 moles

0,575 = [A⁻] / 4,00 moles

2,30 moles = [A⁻]

That means you need 2,30 moles of formate (Sodium formate), to produce the desired buffer. As the molar mass of sodium formate is 68,01g/mol, the weight of sodium formate that must be added is:

2,30mol×(68,01g/mol) = 156,4 g of sodium formate.

I hope it helps!

7 0

3 years ago

I need the answer for the first question 1.

sp2606 [1]

metal salt acid hydroyon and why cuz I guessed and I haven't t learned this yet and I need points

4 0

3 years ago

You have dissolved 10 g sodium oxide in 200 ml water.calculate concentration of the solution

e-lub [12.9K]

Answer:

0.85 Molar Na2O

Explanation:

Determine the moles of sodium oxide, Na2O, in 10 grams by dividing by the molar mass of Na2O (61.98 g/mole).

(10 g Na2O)/(61.98 g/mole) = 0.161 moles Na2O.

Molar is a measure of concentration. It is defined as moles/liter. A 1 M solution contains 1 mole of solute per liter of solvent. [200 ml water = 0.2 Liters water.]

In this case, we have 0.161 moles Na2O in 0.200 L of solvent.

(0.161 moles Na2O)/(0.200 L) = 0.85 Molar Na2O

8 0

2 years ago

Aluminum chloride is added to cesium does a single replacement reaction occur, and if so what is the correct chemical equation f

krek1111 [17]

The answer is C

I hope that helped

5 0

3 years ago

Read 2 more answers