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Misha Larkins [42]
3 years ago
9

Determine the freezing point of a 3.70 m solution of phenol in benzene. Benzene has a freezing point of 5.5°C and a freezing poi

nt depression constant of 5.12°C•kg/mol.
Which equation should you use?
Chemistry
2 answers:
bearhunter [10]3 years ago
7 0

Answer-ΔTf = Kfm

Explan

olganol [36]3 years ago
7 0

all 3 answers for edge

b

18.9

-13.4

Explanation:

You might be interested in
Define atmpospheric pressure
krok68 [10]

Answer:

that pressure is called atmospheric pressure or air pressure. It is the force exerted on a surface by the air above is as gravity pulls it to earth. atmospheric pressure is commonly measured with a barometer. In a barometer , a column of mercury is a glass tube rises ot falls as the weight of the atmospheric changes

5 0
3 years ago
Balloon has a volume of 600-ml at temperature of 360 K. If the temperature of
Molodets [167]

Answer:

V₂ ≈416.7 mL

Explanation:

This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.

  • V₁ / T₁ = V₂ / T₂

where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.

The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,

  • V₁= 600 mL
  • T₁= 360 K
  • T₂= 250 K

Substitute the values into the formula.

  • 600 mL /360 K = V₂ / 250 K

Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.

  • 250 K * 600 mL /360 K = V₂ / 250 K * 250 K
  • 250 K * 600 mL/360 K = V₂

The units of Kelvin cancel, so we are left with the units of mL.

  • 250 * 600 mL/360=V₂
  • 416.666666667 mL= V₂

Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.

  • 416.7 mL ≈V₂

The volume of the balloon at 250 K is approximately 416.7 milliliters.

5 0
2 years ago
Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -
kari74 [83]

Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)  \Delta H=-2220.1kJ/mol

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]

\Delta H_{C_3H_8}=72.4kJ/mol

Therefore, the heat of formation of propane is 72.4 kJ/mol

7 0
3 years ago
The combustion of alkanes, alkenes and alkynes produces:
algol [13]
Alkanes, alkenes, and alkynes are compounds containing only carbon and hydrogen. When these compounds undergo complete combustion, the products are carbon dioxide and water. Otherwise, when incomplete combustion occurs, carbon dioxide, water, and carbon monoxide are produced. Note that carbon monoxide is a toxic gas that is harmful to people's health.
7 0
3 years ago
Gas stoichiometry: what volume of oxygen at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53 grams
photoshop1234 [79]
Answer is: <span>volume of oxygen is 14.7 liters.
</span>Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.<span>
m(</span>C₃H₈-propane) = 5.53 g.
n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).
n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.
n(C₃H₈) = 0.125 mol.
From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.
n(O₂) = 0.625 mol.
T = 25° = 298.15K.
p = 1.04 atm.
<span>R = 0.08206 L·atm/mol·K.
</span>Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.
V(O₂) = 14.7 L.

5 0
3 years ago
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