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Serggg [28]
3 years ago
5

Will mark as Brainliest.

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
4 0
<span>If I done the math correctly it is 3729J because you multiply 16.5 g  by the 2260 J/g and get 3729 J</span>
You might be interested in
6. What causes the phases of the moon?
lawyer [7]

Answer:

The rotation of the Earth.

6 0
2 years ago
A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this
NeTakaya

Answer:

a)  molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane CH_4}= 16 g/mol

Molar mass of ethane C_2H_6= 30 g/mol

Molar mass of ethylene C_2H_4 = 28 g/mol

Molar mass of water H_2O=18g/mol

number of moles = \frac{mass}{molar mass}

Their molar composition can be calculated as follows:

n_{CH_4}= \frac{75}{16}

n_{CH_4}= 4.69 moles

n_{C_2H_6} = \frac{10}{30}

n_{C_2H_6} = 0.33 moles

n_{C_2H_4} = \frac{5}{28}

n_{C_2H_4} = 0.18 moles

n_{H_2O}= \frac{10}{18}

n_{H_2O}= 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = \frac{n_{H_2O}}{n_{drygas}}

= \frac{0.56}{5.2}

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

    CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O

4.69         2× 4.69

moles       moles

   C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O

0.33      3.5 × 0.33

moles    moles

    C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O

0.18           3× 0.18

moles        moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h

Total moles of O_2 required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles O_2

Thus, moles of air required = \frac{1}{0.21}*11.075

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0
3 years ago
Help please! -A compound contains 0.013 moles of carbon, 0.039 moles of hydrogen and 0.0065 moles of oxygen. Determine the empir
stepan [7]

A) C2H6O1

To find the emperical formula, divide each mole value by the smallest

For carbon, 0.013/0.0065 = 2

For hydrogen, 0.038/0.0065= 6

For oxygen, 0.0065/0.0065= 1

Emperical formula = C2H6O1

8 0
3 years ago
Read 2 more answers
Complete the following sentences:
denpristay [2]

Answer:

Explanation:

a) In an exothermic reaction, the energy transferred to the surroundings from forming new bonds is ___more____ than the energy needed to break existing bonds.  

b) In an endothermic reaction, the energy transferred to the surroundings from forming new bonds is ___less____ than the energy needed to break existing bonds.  

c) The energy change of an exothermic reaction has a _____negative_______ sign.  

d) The energy change of an endothermic reaction has a ____positive________ sign.

The energy changes occur during the bonds formation and bonds breaking.

There are two types of reaction endothermic and exothermic reaction.

Endothermic reactions:

The type of reactions in which energy is absorbed are called endothermic reactions.

In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.

For example:

C + H₂O   →  CO  + H₂

ΔH = +131 kj/mol

it can be written as,

C +  H₂O  + 131 kj/mol  →  CO  + H₂

Exothermic reaction:

The type of reactions in which energy is released are called exothermic reactions.

In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.

For example:

Chemical equation:

C + O₂   →  CO₂

ΔH = -393 Kj/mol

it can be written as,

C + O₂   →  CO₂ + 393 Kj/mol

5 0
3 years ago
A 10.0 g sample of an unknown liquid is vaporized at 120.0°C and 5.0 atm. The volume of the vapour is found to be 568.0 mL. The
Minchanka [31]

Answer:

molecular formula of liquid = C₈H₁₈

Explanation:

First we determine the empirical formula of the liquid:

Number of moles of each element present in the liquid = % mass / molar mass

For Carbon, (molar mass = 12.01 g/mol) : 84.2/12.01 =7.011 moles

For Hydrogen (molar mass = 1.01 g/mol) : 15.8/1.01 = 15.643

Simplest mole ratio of the elements, C : H  is given by:

C = 7.011/7.011 = 1.0

H = 15.643/7.011 = 2.23

Multiplying through with 5, C:H = 5:11

Therefore, empirical formula is C₅H₁₁

The molecular mass of the liquid is next determined:

Using PV = nRT to find the number of moles of the liquid present

P = 5.0 atm; V = 568.0 mL = 0.568 L; R = 0.082 L*atmmol⁻¹ K⁻¹; T = 273 + 120 = 393 K

n = PV/RT = (5*0.568)/0.082*393

n = 0.088 moles

Molar mass of liquid = mass/no of moles = 10.0 g/ 0.088 moles = 113.63 gmol⁻¹

Molecular formula = n(empirical formula)

Molar mass of empirical formula, C₅H₁₁ = 71 gmol⁻¹

n = molecular mass/empirical mass = 113.63/71 = 1.6

Therefore, molecular formula =  1.6*(C₅H₁₁) = C₈H₁₈

6 0
3 years ago
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