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stepan [7]
3 years ago
13

A 35.0 ml sample of 0.225 m hbr was titrated with 42.3 ml of koh. What is the concentration of the koh?

Chemistry
1 answer:
Lemur [1.5K]3 years ago
5 0

Answer:

The concentration of KOH is 0.186 M

Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

This is given as;

KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)

From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.

We use the acid base formular in calculating unknown concentrations. This is given as;

\frac{CaVa}{CbVb}  = \frac{na}{nb}

where;

Ca = Concentration of acid

Va = Volume of acid

Cb = Concentration of base

Vb = Volume of base

na = Number of moles of acid

nb = Number of moles of base

KOH is the base and HBr is acid.

Hence;

Ca = 0.225

Va = 35

Cb = ?

Vb = 42.3

na = 1

nb = 1

Making Cb subject of formular we have;

Cb = \frac{CaVaNb}{VbNa}

Cb = (0.225 * 35 * 1) / (42.3 * 1)

Cb = 0.186 M

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Alchen [17]

Answer : The molar heat of solution of KCl is, 17.19 kJ/mol

Explanation :

First we have to calculate the heat of solution.

q=c\times (\Delta T)

where,

q = heat produced = ?

c = specific heat capacity of water = 1.28kJ/K

\Delta T = change in temperature = 0.360 K

Now put all the given values in the above formula, we get:

q=1.28kJ/K\times 0.360K

q=0.4608kJ=460.8J

Now we have to calculate the molar heat solution of KCl.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 460.8 J

m = mass of KCl = 2.00 g

Molar mass of KCl = 74.55 g/mol

\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{2.00g}{74.55g/mole}=0.0268mole

Now put all the given values in the above formula, we get:

\Delta H=\frac{460.8J}{0.0268mole}

\Delta H=17194.029J/mol=17.19kJ/mol

Therefore, the molar heat of solution of KCl is, 17.19 kJ/mol

7 0
2 years ago
The blank is the material that wave move through
miss Akunina [59]

Waves travel through matter, so I am 99.9% sure that is the answer.

Id look through your lesson to be sure!

4 0
3 years ago
Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in
IgorC [24]

Answer:

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

                                       = 99.7°C - 20°C

                                       = 79.7° C

Mass of water is 2000 g ( 2000 mL × 1 g/mL)

Specific heat of water is 4.2 J/g°C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

                      = 669,480 joules

But, 1 kJ = 1000 J

Therefore, heat change is 669.48 kJ

7 0
3 years ago
#2: A gas at unknown pressure and a volume of 66 cm has its pressure changed to 150 kPa. The new volume
Murljashka [212]

Answer:

<h2>377 kPa</h2>

Explanation:

The original pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we're finding the original pressure

P_1 =  \frac{P_2V_2}{V_1}  \\

150 kPa = 150,000 Pa

We have

P_1 =  \frac{150000 \times 166}{66}  =  \frac{24900000}{66}  \\  = 377272.72...

We have the final answer as

<h3>377 kPa</h3>

Hope this helps you

7 0
3 years ago
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stellarik [79]

Answer:

B

Explanation:

Particles in a solid have fixed locations in a volume that does not change. Solids have a definite volume and shape because particles in a solid vibrate around fixed locations.

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