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Mkey [24]
3 years ago
6

the freezing point depression constants of the solvents cyclohexane and naphthalene are 20.1°C/m and 6.94°C/m respectively. Whic

h would give a more accurate determination by freezing point depression of the molar mass of a substance that is soluble in either solvent? Why?
Chemistry
1 answer:
Evgesh-ka [11]3 years ago
7 0

Explanation:

It'd be better to use cyclohexane.  The possible explanation is that the freezing temperature will change by 20.1 degrees for each mole of substance added to 1 kg of cyclohexane, although the same amount added to naphthalene will change its freezing point just by 6.94 degrees.

It is so much easier to identify a larger change more adequately than a smaller one.  You would actually not have a 1 molal solution in operation, so the variations in freezing points would be even smaller than the ones already described.

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Complete combustion of a compound containing hydrogen and carbon produced 2.641 g of carbon dioxide and 1.442 grams of water as
klio [65]

  The empirical   formula  of hydrocarbon is C₃H₈

The  molecular formula of hydrocarbon   is C₆H₁₆


<u><em> Empirical  formula  calculation</em></u>

Hydrocarbon  contain  carbon and hydrogen

  Step 1:  find the  mass  carbon (C) in carbon dioxide (CO₂)  and hydrogen (H )  in water

mass of  of element = molar mass  of element/ molar mass molecule x total mass of    molecule

From periodic table the molar mass  of C =12,    for CO₂ = 12+( 16 x2) =44 g/mol,     for H = 1.00 g/mol,    for H₂O = (2 x1)+16 = 18 g/mol

mass of C = 12/44 x 2.641 =0.7203 g

since there are 2 atom  of H in H₂O the molar mass of H = 1 x2 = 2 g/mol

mass of H  is therefore =  2/18 x 1.442 =0.1602 g


Step 2:  find the moles of C and H

moles = mass÷ molar mass

moles of C = 0.7203 g÷ 12 g/mol = 0.060  moles

moles of H  =  0.1602÷ 1 g/mol = 0.1602 moles


Step 3: find the mole ratio  of C and H by dividing  each  mole by smallest mole ( 0.06)

for C = 0.06/0.06 =1

  For H = 0.1602/0.06 =2.67

multiply   by 3  to remove the decimal

For C = 1 x3 =3

For H = 2.67 x3 =8

therefore the empirical formula = C₃H₈


<u><em>The molecular formula calculation</em></u>

[C₃H₈]n  = 88.1 g/mol

[12 x 3)+( 1 x8)]n =88.1 g/mol

44 n = 88.1

divide both side by 44

n=2

therefore [C₃H₈]₂   = C₆H₁₆



7 0
3 years ago
Read 2 more answers
What data would be most helpful to support the claim that the population of a certain plant is dependent on the rainfall in its
LekaFEV [45]

Answer: number 1 rainfall amounts

Explanation: i hope this heps you bye

5 0
3 years ago
A gas sample enclosed in a rigid metal container at room temperature (20.0∘C) has an absolute pressure p1. The container is imme
Vlad [161]

Answer : The new absolute pressure is, 1.068\times P_1

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

\frac{P_2}{P_1}=\frac{T_2}{T_1}

where,

P_1 = initial pressure of gas

P_2 = final pressure of gas

T_1 = initial temperature of gas = 20.0^oC=273+20.0=293.0K

T_2 = final temperature of gas = 40.0^oC=273+40.0=313.0K

Now put all the given values in the above equation, we get:

\frac{P_2}{P_1}=\frac{313.0K}{293.0K}

\frac{P_2}{P_1}=1.068

P_2=1.068\times P_1

Therefore, the new absolute pressure is, 1.068\times P_1

5 0
4 years ago
Which products are the result of a neutralization reaction? (1 point)
Nata [24]

Answer:

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3 0
3 years ago
A water treatment plant applies chlorine for disinfection so that 10 mg/L chlorine is achieved immediately after mixing. The vol
Vlada [557]

<u>Answer:</u> The mass of chlorine needed by the plant per day is 9.4625\times 10^8mg

<u>Explanation:</u>

We are given:

Volume o water treated per day = 25,000,000 gallons

Converting this volume from gallons to liters, we use the conversion factor:

1 gallon = 3.785 L

So, \frac{3.785L}{1\text{ gallon}}\times 25,000,000\text{ gallons}=9.4625\times 10^7L

Amount of chlorine applied for disinfection = 10 mg/L

Applying unitary method:

For 1 L of water, the amount of chlorine applied is 10 mg

So, for 9.4625\times 10^7L of water, the amount of chlorine applied will be \frac{10mg}{1L\times 9.4625\times 10^7L}=9.4625\times 10^8mg

Hence, the mass of chlorine needed by the plant per day is 9.4625\times 10^8mg

6 0
3 years ago
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