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alukav5142 [94]
3 years ago
15

. In a titration, a 25.0 mL sample of 0.150 M HCl is neutralized with 44.45 mL of Ba(OH)2. a. Write the balanced molecular equat

ion for this reaction. __________________________________________________________ _________________ b. What is the molarity of the base solution
Chemistry
1 answer:
choli [55]3 years ago
5 0

Answer:

Equation of reaction:

a) 2HCl + Ba(OH)2 ==> CaCl2 + 2H2O

b) Molarity of base = 0.042 M.

Explanation:

Using titration equation

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2

NB is the number of mole of base = 1

CA is the molarity of acid =0.15M

CB is the molarity of base = to be calculated

VA is the volume of acid = 25 ml

VB is the volume of base = 44.45mL

Substituting

0.15×25/CB×44.45 = 2/1

Therefore CB =0.15×25×1/44.45×2

CB = 0.042 M.

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7 0
3 years ago
5 points
Orlov [11]

Answer:

The correct option is (a).

Explanation:

We need to find the product of given numbers.

First number = 29.5

Second number = 240

Product of two numbers = 29.5 × 240

= \dfrac{295}{10}\times 240

Zeroes from numerator and denominator get cancelled.

So,

=295\times 24\\\\=7080

Hence, the correct option is (a).

7 0
3 years ago
If a aluminum rod was made out of solid glass instead of solid aluminum what would happen.
Alika [10]

Answer:

it would break when it hits the ground

Explanation:

5 0
3 years ago
An object has a mass of 6.8 g and volume of 34 mL. What is the density of the object?​
jenyasd209 [6]

Answer:

<h2>Density = 0.2 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>Density =  \frac{mass}{volume}</h3>

From the question the points are

mass = 6.8 g

volume = 34 mL

Substitute the values into the above formula and solve

That's

<h3>Density =  \frac{6.8}{34}</h3>

We have the final answer as

<h3>Density = 0.2 g/mL</h3>

Hope this helps you

7 0
3 years ago
A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0
3 years ago
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