Name the following compound: CH3-CH2-CH2-CH2-CH3 CH3 CH3

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Tema [17]

Answer:

Match the words to the definitions.

Explanation:

1. F

2. A

3. C

4. E

5. B

6. D

7. B

8. F

9. A

10. E

11. D

12. C

13. B

4 0

3 years ago

An alkaline battery produces electrical energy according to the following equation.

Gnesinka [82]

Answer:

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Explanation:

Part a

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Limiting reactant = ?

Given Reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

To determine the limiting reactant first we will look at the reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

1 mol 2 mol

Convert moles to mass

molar mass of Zn = 65.4 g/mol

Mass of MnO₂ = 55 + 2 (16) = 55 + 32 = 87 g/mol

So,

Zn(s) + 2 MnO₂(s) + H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)

1 mol (65.4 g/mol) 2 mol (87 g/mol)

65 g 174 g

So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.

now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.

So, for MnO₂ if we calculate the needed amount of zinc

So apply unity formula

65 g Zn react ≅ 174 g of MnO₂

X g of Zn ≅ 21.5 g of MnO₂

Do cross multiplication

X g Zn react = 65 g x 21.5 g / 174 g

X g of Zn ≅ 8.032 g

So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.

So MnO₂ will be consumed completely an it will be limiting reactant.

____________

part b

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Mass of Zn(OH)₂ produced = ?

Given Reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

Solution:

As from the part A we come to know that MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.

So first we convert mass of MnO₂ to moles

Formula Used

no. of moles = mass in grams / molar mass

Mass of MnO₂ = 55 + 2 (16)

Mass of MnO₂ = 55 + 32 = 87 g/mol

Put values in the above equation

no. of moles = 21.5 g / 87 g/mol

no. of moles = 0.25 moles

Now,

Look at the reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

2 mol 1 mol

So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂

then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂

So.

apply unity formula

2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂

0.25 moles of MnO₂ ≅ X mole of Zn(OH)₂

Do cross multiplication

X mole of Zn(OH)₂ = 1 mole x 0.25 mol x / 2 mol

X mole of Zn(OH)₂ ≅ 0.125

Now Conver moles of Zn(OH)₂ to mass

Formula used

mass in grams = no. of moles x molar mass

Molar mass of Zn(OH)₂

Molar mass of Zn(OH)₂ = 65.4 + 2 (16 + 1)

Molar mass of Zn(OH)₂ = 65.4 + 2 (17)

Molar mass of Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol

Put values in above equation

mass in grams = 0.125 mol x 99.4 g/mol

mass in grams = 12.43 g

So,

12.43 g of Zn(OH)₂ will be produce.

5 0

3 years ago

What are some examples of monomers and oligomers?

aleksandr82 [10.1K]

What are some examples of monomers and oligomers?
Organic molecules, such as proteins, carbohydrates, lipids and nucleic acids, are made of simple subunits called monomers. Plasticizers are oligomeric esters widely used to soften thermoplastics such as PVC and urethane acrylate .

If a chemical compound accelerates and regulates metabolic reactions, which type of role does it play - structural or physiological?
I believe the function that it plays would be physiological since it focuses more on the regulation of the reactions inside the body.

7 0

3 years ago

Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +

nikklg [1K]

Answer:

For 1: The value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2: The value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

Explanation:

For the given chemical equation:

$H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)$

For 1:

To calculate the $\Delta G$ for given value of equilibrium constant, we use the relation:

$\Delta G=-RT\ln K_p$ .....(1)

where,

$\Delta G$ = ? kJ/mol

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2:

The expression of $K_p$ for the given chemical equation is:

$K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}$

We are given:

$p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm$

Putting values in above equation, we get:

$K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52$

Now, calculating the value of $\Delta G$ by using equation 1:

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

3 0

3 years ago

Read 2 more answers

Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and

swat32

Answer:

$\%\ mass\ of\ CaCO_3=93.37\ \%$

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (20 + 273.15) K = 293.15 K

T = 293.15 K

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 62.3637 L.mmHg/K.mol

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K

⇒n of $CO_2$ produced = 0.0493 moles

According to the reaction:-

$CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2$

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of $CaCO_3$ = 100.0869 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0493\ mol= \frac{Mass}{100.0869\ g/mol}$

$Mass_{CaCO_3}=4.93\ g$

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

$\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100$

$\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100$

$\%\ mass\ of\ CaCO_3=93.37\ \%$

3 0

3 years ago