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Alex_Xolod [135]
3 years ago
10

KOH and LiOH are examples of bases neutral compounds acids salts

Chemistry
2 answers:
vampirchik [111]3 years ago
8 0
It is an example of bases
svetlana [45]3 years ago
3 0

Answer : The correct option is, Bases.

Explanation :

Acids : It is a substance that has ability of donating proton or hydrogen ion, H^+.

Bases : It is a substance that has ability of donating hydroxide ion, OH^-.

Salts : It is an ionic compound that contain positively charged ions and negatively charged ions. The salt is formed when an acid react with a base in equal proportions.

Neutral compound : It is a compound that is formed by the combination of strong acid and strong base or weak acid and weak base.

As per question, KOH and LiOH are the examples of bases because they have ability of donating hydroxide ion, OH^-. KOH and LiOH are the strong base that are completely dissociate.

KOH\rightarrow K^++OH^-

LiOH\rightarrow Li^++OH^-

Hence, the correct option is, Bases.

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Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
Ions form as a result of uneven charge distribution between the nucleus and the electron cloud. TRUE OR FALSE
storchak [24]

Answer:

true

Explanation:

true

8 0
3 years ago
Which of these is an example of an engineering solution?
agasfer [191]

Answer: Creating a biodegradable trash bag

5 0
4 years ago
Read 2 more answers
Electron configuration of Phosphorus
Aleks [24]
1s^2 2s^2 2p^6 3s^2 2p^3 or the shortcut way is [Ne] 3s^2 2p^3
3 0
3 years ago
Show the calculation of the mole fraction of each gas in a 1.00 liter container holding a mixture of 8.65 g of CO2 and 4.32 g of
Vlad1618 [11]

Answer:

0.745 and 0.245

Explanation:

Mole fraction (χ) is  the number of moles of a component divided by the total number of moles in a mixture.  

We must calculate the moles of each component.  

Let CO₂ be Gas 1 and SO₂ be Gas 2.

1. Calculate the moles of each gas.

\text{n} _{1} = \text{8.65 g} \times \dfrac{\text{1 mol}}{\text{44.01 g}} = \text{0.1965 mol}\\\\\text{n} _{2} = \text{4.32 g} \times \dfrac{\text{1 mol}}{\text{64.06 g}} = \text{0.067 44 mol}

2. Calculate the total moles.

n_{\text{tot}} = \text{n}_{1} + \text{n}_{2} = \text{0.1965 mol} +\text{0.067 44 mol} = \text{0.2640 mol}

3. Calculate the mole fraction of each component

\chi_{1} =\dfrac{0.1965}{0.2640} = \mathbf{0.745}\\\\\chi_{2} =\dfrac{0.06744}{0.2640} = \mathbf{0.255}

6 0
3 years ago
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