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sukhopar [10]
2 years ago
8

A laboratory technician needs to prepare 500 milliliters of a 2 M solution of sodium hydroxide (NaOH) in a flask. The flask has

a volume of exactly 500 milliliters. Which quantity does the laboratory technician need to measure before making the solution?
Chemistry
1 answer:
Aloiza [94]2 years ago
3 0

Answer : The laboratory technician need to measure before making the solution is the mass of the solute.

Explanation :

Given : Molarity of NaOH = 2 M = 2 mole/L

Volume of solution = 500 ml

Molar mass of NaOH = 40 g/mole

For making the solution, we need the mass of NaOH.

Molarity : It is defined as the number of moles of solute present in on liter of solution.

Formula used : M=\frac{w_1\times 1000}{M_1\times V_s}

where,

M = molarity of the NaOH

w_1 = mass of NaOH

M_1 = molar mass of NaOH

V_s = volume of solution

Now put all the given values in this formula, we get the mass of NaOH.

2mole/L=\frac{w_1\times 1000}{(40g/mole)\times (500)L}

w_1=40g

Therefore, the mass of NaOH is 40 gram.

As we know the molarity of solute, volume of solution and molar mass of the solute. Hence, the laboratory technician will need to measure only mass of the solute before making the solution.

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denis23 [38]

Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of CHCl_3 = 1.492 g/mL

The density of CHBr_3 = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

6 0
3 years ago
Write an equilibrium equation that shows bisulfate acting as a weak acid in water.
NemiM [27]

Answer:

Explanation:

Bisulphate ion is a weak acid as it can form hydronium ion in water .

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

The equilibrium constant of this reaction is very small , hence bisulphate ion is very weak acid.

4 0
3 years ago
Water is amphoteric. If it reacts with a compound that is a stronger acid than itself, water acts as a
kirza4 [7]

Answer:

Water acts as a base in the presence of a strong acid

Explanation:

Water,being an amphoteric compound, can act both as an acid and as a base.

In the presence of an acid , water acts as a base but in the presence of a base, water acts as an acid.

5 0
2 years ago
How does the concentration of a gas in solution change when the partial pressure of the gas above the solution increases?
irakobra [83]
Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures
5 0
2 years ago
A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t
Lilit [14]

Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T} =k

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

\frac{P1}{T1} =\frac{P2}{T2}

In this case:

  • P1= 2 atm
  • T1= 50 C= 323 K (being 0 C= 273 K)
  • P2= 3.2 atm
  • T2= ?

Replacing:

\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}

Solving:

T2*\frac{2 atm}{323 K} =3.2 atm

T2=3.2 atm*\frac{323 K}{2 atm}

T2= 516.8 K= 243.8 C

<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>

5 0
3 years ago
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