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3 years ago

1. Calculate the heat change associated with cooling a 350.0 g aluminum bar from

Chemistry

1 answer:

Amanda [17]3 years ago

7 0

Answer:

14175 j heat released.

Explanation:

Given data:

Mass of aluminium = 350.0 g

Initial temperature = 70.0°C

Final temperature = 25.0°C

Specific heat capacity of Aluminium = 0.9 j/g.°C

Heat changed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Heat change:

ΔT = Final temperature - initial temperature

ΔT = 25.0°C - 70°C

ΔT = -45°C

Q = m.c. ΔT

Q = 350 g × 0.9 j/g.°C × -45°C

Q = -14175 j

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3 years ago

In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.

Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂ → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

C₁₀H₈O : C₁₁H₇O₂Cl

1 : 1

1734.1 : 1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

C₁₀H₈O : C₁₁H₇O₂Cl

1 : 1

0.694 : 0.694

COCl₂ : C₁₁H₇O₂Cl

1 : 1

1.0 : 1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield = 143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

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Why is it important to add coefficients and not change the subscripts when balancing a chemical equation?

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You dissolve 8.65 grams of lead(l) nitrate in water and then you add 2 50 grams of aluminum. This reaction occurs 2AI(S)+ 3Pb(NO

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Answer: The theoretical yield of solid lead comes out to be 5.408 grams.

Explanation:

To calculate the moles, we use the following equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of Lead nitrate:

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles$

Moles of Aluminium:

Given mass of aluminium = 2.5 grams

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{2.5g}{27g/mol}=0.0925moles$

For the given chemical reaction, the equation follows:

$2AI(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2AI(NO3)_3(aq.$

By Stoichiometry:

3 moles of lead nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead nitrate are produced by = $\frac{2}{3}\times 0.0261=0.0174moles$ of aluminium.

As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.

Lead nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of lead nitrate are produces 3 moles of lead metal.

So, 0.0261 moles of lead nitrate will produce = $\frac{3}{3}\times 0.0261=0.0261moles$ of lead metal.

Now, to calculate the grams or theoretical yield of lead metal, we put in the mole's equation, we get:

Molar mass of lead = 207.2 g/mol

Putting values in above equation, we get:

$0.0261mol=\frac{\text{Given mass}}{207.2g/mol}$

Mass of lead = 5.408 grams

Hence, the theoretical yield of solid lead comes out to be 5.408 grams.

8 0

3 years ago