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1 year ago

8

you are give a bottle that contains 4.59cm3 of a metalic solid .the total mass of the bottle plus solid =35.66g .the empty bottl

e weighs 14.23g .what is the density of the solid

1 answer:

kirill115 [55]1 year ago

6 0

Answer:

Density = 4.67 g/cm³

Explanation:

Density can be found using the following formula:

Density = mass (grams) / volume (cm³)

The mass of the solid can be found by subtracting the weight of the empty bottle from the weight of the bottle + solid.

35.66 g - 14.23 g = 21.43 g solid

Now that you have the volume and mass, you can determine the density via the formula.

Density = mass / volume

Density = 21.43 g / 4.59 cm³

Density = 4.67 g/cm³

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D) blue

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3 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

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Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

Generally the concentration of base

$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

Now after the reaction the number of moles of base is zero i.e has been used up

this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

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Answer:

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Answer:

C.

Explanation:

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