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3 years ago

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Chemistry

1 answer:

mestny [16]3 years ago

4 0

Answer:

Uses of Sodium chloride:

1.it is used as table salt in our diets

2. it is used as a food preservative

3. It is used as an industrial source of sodium and chlorine

Uses of Sodium hydroxide:

1. It is in the laboratory as an alkali in the neutralization of acids

2. It is used in the manufacture of soap

3. It is used in the refining of petroleum

Uses of Sodium carbonate:

1. It is used in the manufacture of detergents and glass

2. It is used to soften hard water

3. it is used to standardize acids in the laboratory

Uses of Sodium sulphate

1. It is used in the manufacture of detergents and glass

2. It is used as a purgative

3. It is used to treat wood-pulp for paper making.

Uses of Potassium nitrate

1. It is used in making gunpowder

2. It is used as rocket propellant

3. It is used in making fertilizer

Uses of Potassium chlorate

1. It is used as an oxidizer

2. It is used in making explosives

3. it is used as a disinfectant

Uses of Potassium permanganate (vii)

1. It is used as an oxidizing agent

2. It is used as a disinfectant

3. It is used as an analytical reagent

Uses of Potassium chloride

1. It is used for the manufacture of fertilizer

2. It is used as an alternative salt to sodium chloride

3. It is used in medicine to treat low blood pressure

2. Potassium nitrate and potassium chloride are used in the manufacture of fertilizers.

Potassium nitrate is used in fertilizers because it provides two important plant nutrients, nitrogen and potassium to plants.

Potassium chloride (potash) is used in fertilizers because it efficiently deliver the important plant nutrient, potassium, to plants.

Explanation:

Uses of Sodium chloride:

1. it is used as table salt in our diets

2. it is used as a food preservative

3. It is used as an industrial source of sodium and chlorine

Uses of Sodium hydroxide:

1. It is in the laboratory as an alkali in the neutralization of acids

2. It is used in the manufacture of soap

3. It is used in the refining of petroleum

Uses of Sodium carbonate:

1. It is used in the manufacture of detergents and glass

2. It is used to soften hard water

3. it is used to standardize acids in the laboratory

Uses of Sodium sulphate

1. It is used in the manufacture of detergents and glass

2. It is used as a purgative

3. It is used to treat wood-pulp for paper making.

Uses of Potassium nitrate

1. It is used in making gunpowder

2. It is used as rocket propellant

3. It is used in making fertilizer

Uses of Potassium chlorate

1. It is used as an oxidizer

2. It is used in making explosives

3. it is used as a disinfectant

Uses of Potassium permanganate (vii)

1. It is used as an oxidizing agent

2. It is used as a disinfectant

3. It is used as an analytical reagent

Uses of Potassium chloride

1. It is used for the manufacture of fertilizer

2. It is used as an alternative salt to sodium chloride

3. It is used in medicine to treat low blood pressure

2. Potassium nitrate and potassium chloride are used in the manufacture of fertilizers.

Potassium nitrate is used in fertilizers because it provides two important plant nutrients, nitrogen and potassium to plants.

Potassium chloride (potash) is used in fertilizers because it efficiently deliver the important plant nutrient, potassium, to plants.

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Acid-base reaction will give salt and water.

The 2nd option is correct.

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Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container

Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

Explanation:

Here Pressure in a container is given as

$P=\frac{1}{3} \rho$

Here

P is the pressure which is to be calculated

ρ is the density of the gas which is to be calculated as below

$\rho =\frac{mass}{Volume of container}$

Here

mass is to be calculated for 75 gas phase molecules as

$m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g$

Volume of container is 0.5 lts

So density is given as

$\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3$

is the mean squared velocity which is given as

$=RMS^2$

Here RMS is the Root Mean Square speed given as 605 m/s so

$=RMS^2\\=(605)^2\\=366025$

Substituting the values in the equation and solving

$P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa$

So the average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

6 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

3 years ago

The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?

Vitek1552 [10]

The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x

We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10

7 0

3 years ago

.........m.......................i need ur help guys​

.........m.......................i need ur help guys