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3 years ago

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HURRY What is true for the energy involved in an explosion? A There is more energy at the end of the explosion than there is at

the beginning. B There is less energy at the end of the explosion than there is at the beginning. C There is the same amount of energy at the end of the explosion than there is at the beginning. D The amount of energy before and after the explosion depends on the type of reaction.

1 answer:

andreyandreev [35.5K]3 years ago

5 0

B .........................................

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How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Cel

o-na [289]

1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa). V = volume, in $m^3$ .

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:

PV= nRT

Given data:

P=100.0 kPa =0.986923 atm

T=100 degree celcius= 100 + 273 =373 K

V=35.5 L

Substituting the values in the equation.

n= $\frac{\;0,98 \;atm \;X \;35,5 \;L }{\;0,082\;atm / \;K mol \;X \;373 K}$

n= 1.137448506 mol

Hence, 1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

Learn more about ideal gas here:

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7 0

2 years ago

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c

gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= <u>135,583 J</u> = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

<u> Now that we have the heat of combustion, we need to calculate the molar heat. </u>

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write the <u>conversion factor as 135.58 kJ/ 4.40 g</u>.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

3 0

4 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0

3 years ago

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How many neutrons does beryllium have?

PilotLPTM [1.2K]

5.

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6 0

3 years ago

A mixture of ethyl acetate vapour and air has a relative saturation of 50% at 303 K and a total pressure of 100 kPa. If the vapo

AURORKA [14]

Answer : The correct option is, (b) 0.087

Explanation :

The formula used for relative saturation is:

$\text{Relative saturation}=\frac{P_A}{P_A^o}$

where,

$P_A$ = partial pressure of ethyl acetate

$P_A^o$ = vapor pressure of ethyl acetate

Given:

Relative saturation = 50 % = 0.5

Vapor pressure of ethyl acetate = 16 kPa

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

$0.5=\frac{P_A}{16kPa}$

$P_A=8kPa$

Now we have to calculate the molar saturation.

The formula used for molar saturation is:

$\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}$

and,

P(vapor free) = Total pressure - Vapor pressure

P(vapor) = $P_A$ = 8 kPa

So,

P(vapor free) = 100 kPa - 8 kPa = 92 kPa

The molar saturation will be:

$\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}$

$\text{Molar saturation}=\frac{8kPa}{92kPa}=0.087$

Therefore, the molar saturation is 0.087

5 0

4 years ago