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EleoNora [17]
3 years ago
7

The table below shows the time that the Moon rose and set over a few days.

Chemistry
1 answer:
Llana [10]3 years ago
4 0

Answer:

C.

rise at 6:20 a.m., set at 4:16 p.m.

Explanation:

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What is the freezing point (in degrees Celcius) of 4.14 kg of water if it contains 235.1 g of butanol, C 4 H 9 O H
Karolina [17]

Answer:

Explanation:

Molal freezing point depression constant of butanol Kf = 8.37⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

mol weight of butanol = 70 g

235.1 g of butanol = 235.1 / 70 = 3.3585 moles

3.3585 moles of butanol dissolved in 4.14 kg of water .

ΔTf = 8.37 x 3.3585 / 4.14

= 6.79⁰C

Depression in freezing point = 6.79

freezing point of solution = - 6.79⁰C .

5 0
2 years ago
Where is most of an atom's atomic mass?
Kamila [148]

Answer:

The nucleus contains the majority of an atom's mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom's volume. The diameter of an atom is on the order of 10−10 m, whereas the diameter of the nucleus is roughly 10−15 m—about 100,000 times smaller.

Explanation:

3 0
2 years ago
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Any two difference between short sightedness and long sightednes​
SSSSS [86.1K]

Explanation:

tala it is also called myopia lekhnu hai

Äni Arko MA it is also called hypermetropia .

3 0
2 years ago
When you drain the solution from the bottom, the concentration will:
djverab [1.8K]

.Answer:

stay the same the conc. of a solution is the same throughout the liquid

Explanation:

3 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m
olga nikolaevna [1]

Explanation:

The reaction equation for the given reaction will be as follows.

               HCl + NaOH \rightarrow NaCl + H_{2}O

Each mole of both HCl and NaOH gives one mole of water.

Also, it is given that 1 liter of NaOH and HCl solution contains 1.6 mol dm^{-3} of NaOH (HCl).

It is known that 1 cm^{3} = 0.001 liter. So, 87 cm^{3} = 0.087 liter.

Hence, number of moles of water obtained from the given reaction are as follows.

                  0.087 liter × 1.6 mol = 0.1392 moles

            No. of moles = \frac{mass}{molar mass of water}

                0.1392 moles = \frac{mass}{18 g/mol}

                       mass = 2.5056 g

Now, volume of water present before the reaction is 2 \times 0.087 liter = 0.174 liter or 0.174 Kg (as density is 1 Kg/cm^{3}) or 174 g (as 1 kg = 1000 g).

Therefore, total weight of water present = 2.5056 g + 174 g = 176.5056 g

Formula to calculate enthalpy of neutralization is as follows.

                Enthalpy of neutralization = mS \Delta T

[/tex]

where,            m = mass

                       S = specific heat capacity

                   \Delta T = change in temperature

Putting the given values in the formula as follows.

       Enthalpy of neutralization = mS \Delta T

                     = 176.5056 g \times 4.18 J/K g \times (317.4 K - 298 K)              

                                                           = 14333.73 J

   or,                                                    = 14.33 kJ

Thus, we can conclude that the enthalpy of neutralization of given reaction is 14.33 kJ.

5 0
3 years ago
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