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2 years ago

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What volume of benzene (C6H6, d= 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to t

he following reaction?2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ

1 answer:

Anton [14]2 years ago

4 0

Answer:

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

Explanation:

<u>Step 1:</u> Data given

Density of benzene = 0.88 g/mL

Molar mass of benzene = 78.11 g/mol

Heat produced = 1.5 * 10³ kJ

<u>Step 2:</u> The balanced equation

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔH°rxn = -6278 kJ

<u>Step 3:</u> Calculate moles of benzen

1.5 * 10³ kJ * (2 mol C6H6 / 6278 kJ) = 0.478 mol C6H6

<u>Step 4:</u> Calculate mass of benzene

Mass benzene : moles benzene * molar mass benzene

Mass benzene= 0.478 * 78.11 g

Mass of benzene = 37.34 grams

<u>Step 5:</u> Calculate volume of benzene

Volume benzene = mass / density

Volume benzene = 37.34 grams / 0.88g/mL

Volume benzene = 42.4 mL

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

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

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

2 years ago

Calculate the amount of heat needed to boil of benzene (), beginning from a temperature of . Be sure your answer has a unit symb

Pavlova-9 [17]

Answer:

Check explanation

Explanation:

From the question, the parameters given are 64.7g of benzene,C6H6; a starting temperature of 41.9°C and bringing it to 33.2°C.

Molar mass of benzene,C6H6= 78.11236 g/mol.

Things to know: heat capacity of benzene, C6H6= 1.63 J/g.K, the heat of fusion = 9.87 kj/mol.

STEP ONE(1): ENERGY USED IN MELTING BENZENE SOLID.

Using the formula below;

Energy used in melting the solid(in JOULES) = (mass of benzene/molar mass of benzene) × heat of Fusion.

=(64.7 g of C6H6/ 78.11236(g per mol) of C6H6) × 9.87 kJ per mol.

= 8.175 J.

= 0.008175 kJ.

STEP TWO (2): ENERGY OF HEATING THE LIQUID.

It can be calculated from the formula below;

Energy= heat capacity (J/g.K) × mass of benzene× (∆T).

= 1.63 J/g.K × 64.7 × (41.9-33.2).

= 917.5J.

= 0.9175 kJ.

Energy required to boil benzene= Energy required to melt the bezene + energy required for boiling.

= 0.008175+ 0.9175.

= 0.93kJ

Approximately, 1 kJ

8 0

2 years ago

How many cm 3 are in 0.014 in 3? (1 in = 2.54 cm)

pashok25 [27]

Answer:

0.229 cm³.

Explanation:

The following data were obtained from the question:

Volume (in in³) = 0.014 in³

Volume (in cm³) =?

1 in = 2.54 cm

Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:

1 in = 2.54 cm

Therefore,

1 in³ = 2.54³ cm³

1 in³ = 16.387 cm³

Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:

1 in³ = 16.387 cm³

Therefore,

0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³

0.014 in³ = 0.229 cm³

Thus, 0.014 in³ is equivalent to 0.229 cm³.

7 0

2 years ago

Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate

Novosadov [1.4K]

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of $K_3PO_4$ = 150 g

Mass of $Al_2(CO_3)_3$ = 90 g

Molar mass of $K_3PO_4$ = 212.27 g/mole

Molar mass of $Al_2(CO_3)_3$ = 233.99 g/mole

Molar mass of $K_2CO_3$ = 138.205 g/mole

First we have to calculate the moles of $K_3PO_4$ and $Al_2(CO_3)_3$

$\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=\frac{150g}{212.27g/mole}=0.7066moles$

$\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=\frac{90g}{233.99g/mole}=0.3846moles$

The given balanced reaction is,

$2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4$

From the given reaction, we conclude that

2 moles of $K_3PO_4$ react with 1 mole of $Al_2(CO_3)_3$

0.7066 moles of $K_3PO_4$ react with $\frac{1}{2}\times 0.7066=0.3533$ moles of $Al_2(CO_3)_3$

But the moles of $Al_2(CO_3)_3$ is, 0.3846 moles.

So, $Al_2(CO_3)_3$ is an excess reagent and $K_3PO_4$ is a limiting reagent.

Now we have to calculate the moles of $K_2CO_3$ .

As, 2 moles of $K_3PO_4$ react to give 3 moles of $K_2CO_3$

So, 0.7066 moles of $K_3PO_4$ react to give $\frac{3}{2}\times 0.7066=1.0599$ moles of $K_2CO_3$

Now we have to calculate the mass of $K_2CO_3$ .

$\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3$

$\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g$

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

$\% yield=\frac{\text{ Theoretical yield}}{\text{ Experimental yield}}\times 100$

$\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%$

Therefore, the % yield of potassium carbonate is, 85.33%

6 0

2 years ago

PLEASE HELPPP WILL GIVE BRAINLIEST!!!

Neporo4naja [7]

Answer:

placental mammals

Explanation:

7 0

2 years ago

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