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Sliva [168]
2 years ago
9

How many grams of soda ash would be needed to produce 1.00 kg of sodium bicarbonate?

Chemistry
1 answer:
GaryK [48]2 years ago
5 0

Answer:

736.1 g of soda ash

Explanation:

Data given:

Amount of sodium bicarbonate = 1 Kg

convert Kg to g

1 Kg = 1000 g

Amount of soda ash needed = ?

Solution:

First we look for the reaction by which soda ash gives sodium bicarbonate.

Chemical formulas

c NaHCO₃

Soda Ash = Na₂CO₃

So the reaction is given below:

                    Na₂CO₃ + H₂O + CO₂ ------------> 2 NaHCO₃

So,

now look for the amount of soda ash and NaHCO₃

                   Na₂CO₃ + H₂O + CO₂ ------------> 2 NaHCO₃

                     1 mole                                           2 mole

As,

one mole of Na₂CO₃ (soda ash) gives 2 moles of sodium bicarbonate (NaHCO₃).

Now convert moles to mass

For which molar masses are given below

Molar mass of Na₂CO₃ = 106 g/mol

Molar mass of NaHCO₃ = 72 g/mol

So,

         Na₂CO₃      +       H₂O     +     CO₂      ------------>    2 NaHCO₃

   1 mole (106 g/mol)                                                    2 mole (72 g/mol)

          106 g                                                                            144 g

Apply unity formula

            106 g of Na₂CO₃ ≅ 144 grams NaHCO₃

                X g of Na₂CO₃ ≅ 1000 grams NaHCO₃

Do cross multiplication

              X g of Na₂CO₃ = 106 g x 1000 g / 144 g

               X g of Na₂CO₃ = 736.1 g

So, 736.1 g of soda ash needed to produce 1000 g or 1 kg of sodium bicarbonate.

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6 0
2 years ago
The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at
antoniya [11.8K]

Answer:

- 7.48

Explanation:

Given:

Concentration of the sugar solution, C = 0.3 M

Temperature, T = 27° C = 273 + 27 = 300 K

Now,

The solute potential is given as:

solute potential = - iCRT

where,

i is the number of particles the particular molecule will make in water

i = 1 for sugar

R is the universal gas constant = 0.0831 liter bar/mole-K

on substituting the respective values, we get

solute potential = - 1 × 0.3 × 0.0831 × 300

or

The solute potential = - 7.479 ≈ - 7.48

8 0
2 years ago
Given that 25.0 mL of mercury has a mass of 340.0 g, calculate (a) the density of mercury and (b) the mass of 120.0 mL of mercur
natita [175]

Answer :

(a) The density of mercury is, 13.6 g/ml

(b) The mass of 120.0 ml of mercury is, 1632 grams

Explanation :

(a) Now we have to calculate the density of mercury.

<u>Given :</u>

Volume of mercury = 25.0 ml

Mass of mercury = 340.0 g

Formula used :

\text{Density of mercury}=\frac{\text{Mass of mercury}}{\text{Volume of mercury}}

\text{Density of mercury}=\frac{340.0g}{25.0ml}=13.6g/ml

Therefore, the density of mercury is, 13.6 g/ml

(b) Now we have to calculate the mass of 120.0 ml of mercury.

As, 25.0 ml of mercury has mass = 340.0 g

So, 120.0 ml of mercury has mass = \frac{120.0ml}{25.0ml}\times 340.0g=1632g

Therefore, the mass of 120.0 ml of mercury is, 1632 grams

3 0
2 years ago
Which factors are needed for organisms to live earth
meriva

Answer: sunlight, water, air, habitat, and food.

Explanation: we are all living organisms and we all have our five basic necessities for survival; sunlight, water, air, habitat, and food.

3 0
2 years ago
a weather balloon is filled with 200L of helium at 27 degree Celsius and 0.950 atm. What would be the volume of the gas at -10 d
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<u>Answer:</u> The volume when the pressure and temperature has changed is 1332.53 L

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=0.950atm\\V_1=200L\\T_1=27^oC=[273+27]K=300K\\P_2=0.125atm\\V_2=?L\\T_2=-10^oC=[273-10]K=263K

Putting values in above equation, we get:

\frac{0.950atm\times 200L}{300K}=\frac{0.125\times V_2}{263K}\\\\V_2=1332.53L

Hence, the volume when the pressure and temperature has changed is 1332.53 L

5 0
2 years ago
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