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Vladimir79 [104]
4 years ago
15

You evaporate all of the water from 100 mL of NaCl solution and obtain 11.3 grams of NaCl. What was the molarity of the NaCl sol

ution?
Chemistry
1 answer:
kicyunya [14]4 years ago
3 0

Answer:

                      Molarity  =  1.93 mol.L⁻¹

Explanation:

             Molarity is the unit of concentration used to specify the amount of solute in given amount of solution. It is expressed as,

                         Molarity  =  Moles / Volume of Solution    ----- (1)

Data Given;

                  Mass  =  11.3 g

                  Volume  =  100 mL  =  0.10 L

First calculate Moles for given mass as,

                   Moles  =  Mass / M.mass

                   Moles  =  11.3 g / 58.44 g.mol⁻¹

                   Moles  =  0.1933 mol

Now, putting value of Moles and Volume in eq. 1,

                        Molarity  =  0.1933 mol ÷ 0.10 L

                        Molarity  =  1.93 mol.L⁻¹

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A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and
zlopas [31]

Answer:

37.1°C.

Explanation:

  • Firstly, we need to calculate the amount of heat (Q) released through this reaction:

<em>∵ ΔHsoln = Q/n</em>

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

  • We can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.

<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>

6 0
3 years ago
How much energy is required to raise the temperature of 3 kg of iron from 20° C to 25°C? Use the table below and this equation:
ElenaW [278]

The energy required to raise the temperature of 3 kg of iron from 20° C to 25°C is 6,750 J( Option B)

<u>Explanation:</u>

Given:

Specific Heat capacity of Iron= 0.450 J/ g °C

To Find:

Required Energy to raise the Temperature

Formula:

Amount of energy required is given by the formula,

Q = mC (ΔT)

Solution:

M = mass of the iron in g

So 3 kg = 3000 g

C = specific heat of iron = 0.450 J/ g °C [ from the given table]

ΔT = change in temperature = 25° C - 20°C = 5°C

Plugin the values, we will get,

Q = 3000 g ×  0.450 J/ g °C ×  5°C

 = 6,750 J

So the energy required is 6,750 J.

7 0
4 years ago
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Sergio039 [100]

Answer:

cell's

Explanation:

3 0
3 years ago
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When is the Sun directly overhead in the continental United States?
astra-53 [7]

Answer:

C.

Explanation:

The sun is directly overhead at noon on the equator on the first day of spring, and on the first day of fall. You would have to be less than 23.5 degrees above or below the equator to have the Sun pass directly overhead. Therefore, it never occurs in the continental US.

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MgSO4equilibrate the hydrolysis reaction
maksim [4K]

Answer:

just guessing, but hydrolisis indicates ; hydro= h2o. lisis= destruction.

Explanation:

in my training, this wud mean destruction of water, or loss of water in the human body.

2 equalibrate, takes sodium chloride, to retain fluids by IV.

3 0
4 years ago
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