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sashaice [31]
3 years ago
5

What is the freezing point of a solution containing 4.78 grams naphthalene (molar mass = 128.2 g/mol) dissolved in 32.0 grams pa

radichlorobenzene?
Chemistry
1 answer:
tester [92]3 years ago
4 0
This is a freezing point depression problem, so it will use the equation:

ΔT = i Kf<span> m
</span>
i = 1 (naphthalene does not dissociate further when dissolved)
Kf = 7.10 C/m (a constant for paradichlorobenzene, which you'd be given)
m = moles of solute / kg of solvent; moles of solute = 4.78 / 128.2 = 0.0373; kg of solvent = 0.032; m = 0.0373 / 0.032 = <span>1.166m

</span>ΔT = 1 x 7.10 x 1.166 = 8.279 C

This means the normal freezing point of pure paradichlorobenzene is decreased by 8.279 C; the normal freezing point (again, something you'd be given) is 53.5 C, so the new freezing point would be 53.5 - 8.279 = 45.221 C.
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one part of dalton's atomic theory states: compounds are formed by combinations of atoms from two or more elements. is this stat
ANEK [815]

Answer:

The statement is considered to be true

Explanation:

The statement is true because when elements chemically combine, there are interactions between their valence electrons, causing the two elements to be bonded together to form what is known as a compound.

Compounds can only be formed from interactions between two or more elements. examples include:

Hydrogen + Oxygen = H2O (water)

Sodium + Chlorine = NaCl

Note that if atoms of the same element combine, what is formed is a molecule, not a compound. Some atoms usually do this to attain stability. examples include = O2 H2 and  N2. They are oxygen molecule, hydrogen molecule, and nitrogen molecule respectively.

Compounds are only formed when different elements combine to attain electronic stability.

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3 years ago
3. Complete the following table.
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Li and CI or C and H
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3 years ago
A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the soluti
lilavasa [31]

<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg

<em>Molar concentration </em>

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

<em>c</em> = 1.14 mol/1 L = 1.14 mol/L

<em>Molal concentration</em>

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg

5 0
3 years ago
Read 2 more answers
Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe
Murljashka [212]

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

  • Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 \frac{J}{g*C} and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 \frac{J}{g*C} *20 C and solving: Q=420 J

  • Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}= 3.333 kJ= 3,333 J (being kJ=1,000 J)

  • Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 \frac{J}{g*C} and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 \frac{J}{g*C} *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

<u><em> The amount of heat to absorb is 6,261 J</em></u>

<u><em></em></u>

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