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ValentinkaMS [17]

3 years ago

5

Show how the following are formed

1 answer:

bogdanovich [222]3 years ago

5 0

I think that beryllium is formed in alcohol I think

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Igneous rock___to form magma a)melts b)weathers c)compacts d)crystallizes NEED HELP ASAP

dangina [55]

Answer:

should be A)Melts.Because magma is liquid i think

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3 years ago

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How many grams of CaCl2 are in .48 moles?

Ilia_Sergeevich [38]

Answer:

110.984 grams

Explanation:

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An ideal gas at constant temperature is confined in an airtight cylinder, and a piston pushes down on the gas in the cylinder. a

FrozenT [24]

Pushing down a piston in an airtight cylinder would tend to increase the pressure in the system since the particle collision and collisions with the walls are now more frequent. Also, pushing further, the gas would undergo a phase change from gas to a liquid.

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3 years ago

I need help on #6 please

kobusy [5.1K]

6. a) Team 3 (b/c their numbers were closer apart then the rest)
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3 years ago

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles

alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of $HF$ .

$\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}$

$\text{Moles of HF}=0.250M\times 1.50L=0.375mol$

Now we have to calculate the value of $pK_a$ .

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

$pK_a=-\log (K_a)$

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

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

The reaction will be:

$HF+OH^-\rightleftharpoons F^-+H_2O$

Initial moles 0.375 0.100 0.375

At eqm. (0.375-0.100) 0 (0.375+0.100)

= 0.275 = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[F^-]}{[HF]}$

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

$pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]$

$pH=3.41$

Thus, the pH of the solution is, 3.41

8 0

3 years ago