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kakasveta [241]
2 years ago
8

When C2H5Cl(g) is burned in oxygen, chlorine gas is produced in addition to carbon dioxide and water vapor. 5145 kJ of heat are

released for every four moles of C2H5Cl(g) burned. Which of the following correctly represents the thermochemical equation?
Chemistry
1 answer:
Arisa [49]2 years ago
7 0

<u>Answer:</u> The chemical equation is written below.

<u>Explanation:</u>

Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

\text{hydrocarbon}+O_2\rightarrow CO_2+H_2O

The chemical equation for the combustion of ethyl chloride follows:

4C_2H_5Cl+13O_2\rightarrow 2Cl_2+8CO_2+10H_2O

We are given:

When 4 moles of ethyl chloride is burnt, 5145 kJ of heat is released.

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

A+\text{heat}\rightleftharpoons B

For an exothermic reaction, heat is getting released during a chemical reaction and is written on the product side

A\rightleftharpoons B+\text{heat}

So, the chemical equation follows:

4C_2H_5Cl+13O_2\rightarrow 2Cl_2+8CO_2+10H_2O+5145kJ

Hence, the chemical equation is written above.

You might be interested in
What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH3NH2, with 20.00 mL of 0.10 M methylammonium c
olya-2409 [2.1K]

Answer:

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

pOH=pKb+log\frac{[salt]}{[base]}

pH = 14- pOH

Let us calculate pOHpOH=3.43+ (-0.397)=3.03

pH=14-pOH=14-3.03=10.97

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M

[base] = [Methylamine]=0.10

After mixing with salt

[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M

pKb= -log[Kb]= 3.43

Putting values

pOH = 3.43+log(\frac{[0.0286]}{0.0714}

4 0
2 years ago
When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro
DanielleElmas [232]

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density  * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

5 0
2 years ago
Element X is a radioactive isotope such that every 82 years, its mass decreases by half. Given that the initial mass of a sample
lesantik [10]

Answer: 17 years

Explanation:

Expression for rate law for first order kinetics  for radioactive substance is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{82years}=8.4\times 10^{-3}years^{-1}

b) for 8900 g of the mass of the sample to reach  7700 grams

t=\frac{2.303}{8.4\times 10^{-3}}\log\frac{8900}{7700}

t=17years

Thus it will take 17 years

4 0
2 years ago
To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if
WARRIOR [948]

Answer:

20.2 g silver placed in 21.6 mL of water and 20.2 g copper placed in 21.6 mL of water

Explanation:

Since the density formula is mass divided volume, you can calculate the density of each of the components of the group and see which of all the groups are equal

In the group

density Ag=20.2 g silver/21.6ml H2O=0.9352g/ml

density Cu=20.2 g copper/21.6ml H2O=0.9352 g/ml

6 0
2 years ago
Read 2 more answers
A 8.6 g sample of methane and 15.6 g sample of oxygen react according to the reaction in the video. identify the limiting reacta
GalinKa [24]
Answer:

<span>23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2. </span>

Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O

3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol

4) Convert the reactant masses to number of moles, using the formula 

number of moles = mass in grams / molar mass


CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
<span />

O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂

Which is what the first part of the answer says.

6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.

Which is what the second part of the answer says.

7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.

Which is what the chosen answer says.

8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.

3 0
2 years ago
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