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ArbitrLikvidat [17]
3 years ago
12

How many moles of each substance are needed to prepare the following solutions?

Chemistry
1 answer:
Afina-wow [57]3 years ago
4 0
A) 7.5 % m / V 

7.5 g -------- 100 mL ( solution )
mass g ------ 45.0 mL

mass g = 45.0 x 7.5 / 100

mass g = 337.5 / 100 => 3.375 g

1 mole KCl -------------- 74.55 g
moles KCl ---------------- 3.375 g

moles KCl = 3.375 x 1 / 74.55 => 0.0452 moles of KCl
______________________________________________

b) 9.5 % ( m / V ) :

9.5 g -------- 100 mL ( solution )
mass g ------ 400.0 mL

mass g = 400.0 x 9.5 / 100

mass g = 3800 / 100 => 38 g

1 mole acetic acid -------------- 60.05 g
moles acetid acid  ---------------- 38 g

moles acetid acid = 38 x 1 / 60.05 => 0.6328 moles of  acetid acid<span> 
</span>______________________________________________________

hope this helps!


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A sample of gas X occupies 10 mº at a pressure of 120 kPa.
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Answer:

The new pressure of the gas comes out to be 400 KPa.

Explanation:

Initial volume of gas = V = 10\textrm{ m}^{3}

Initial pressure of gas = P = 120 KPa

Final volume of gas = V' = 3\textrm{ m}^{3}

Assuming temperature to be kept constant.

Assuming final pressure of the gas to be P' KPa

PV = P'V' \\120\textrm{ KPa}\times 10\textrm{ m}^{3} = \textrm{P'}\times 3\textrm{ m}^{3} \\\textrm{P'} = 400\textrm{ KPa}

New pressure of gas = 400 KPa

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4 years ago
25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?
weeeeeb [17]

Answer:

pH = 11.95≈12

Explanation:

Remember  the reaction among aqueous acetic acid (CH_3COOH) and aqueous sodium hydroxide (NaOH)

CH_3COOH + NaOH ->  CH_3COONa + H_2O

First step. Need to know how much moles of the substances are present

0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH = 0.0025 mol NaOH

0.003 mol NaOH * 1 mol CH_3COOH/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH -> Na^{+} + OH^{-}

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95

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