270kJ are needed to keep a 75 watt light bulb burning for 1 hour. How many kilocalories are needed to burn this light bulb for 3 hours?

Answer:

Option C. 80.5%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Next, we shall determine the mass of C3H8 that reacted and the mass of CO2 produced from the balanced equation.

This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44 g/mol

Mass of C3H8 from the balanced equation = 1 x 44 = 44 g

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Mass of CO2 from the balanced equation = 3 x 44 = 132 g

From the balanced equation above,

44 g of C3H8 reacted to produce 132 g of CO2.

Next, we shall determine the theoretical yield of CO2.

This can be obtained as shown below:

From the balanced equation above,

44 g of C3H8 reacted to produce 132 g of CO2.

Therefore, 0.295 kg (i.e 295 g) will react to produce = (295 x 132)/44 = 885 g of CO2.

Therefore, the theoretical yield of CO2 is 885 g.

Finally, we shall determine the percentage yield of CO2 as follow:

Actual yield of CO2 = 712 g

Theoretical yield of CO2 = 885 g

Percentage yield of CO2 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield of CO2 = 712/885 x 100

Percentage yield = 80.5%

Therefore, the percentage yield of CO2 is 80.5%.

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

• The number of moles present in each element is first determined:
• Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

• Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

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The chemical equation presented in option A follows the law of conservation of mass.

The principle of conservation of mass states, mass can neither be created nor destroyed but can be transformed from one form to another.

A reaction that follows the law of conservation of mass,  must have equal number of moles each elements in reactants side and products side.

Only option A follows the law of conservation of mass;

Thus, we can conclude that the chemical equation presented in option A follows the law of conservation of mass.

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Answer:

Option A

Explanation:

The expression that obeys the law of conservation of mass is choice A;

         2LiOH  +  H₂CO₃   →   Li₂CO₃  + 2H₂O

According to the law of conservation of mass; "in a chemical reaction, matter is neither created nor destroyed". By this law, mass is usually conserved.

The equation shows that mass is conserved because the number of moles of each specie is found on both sides

                                                Number of moles

                                 Li                   O                  H              C

Reactants                 2                    5                   4               1

Products                   2                    5                   4               1

This shows that mass is indeed conserved.

Answer:

The law is observed in the given equation.

Explanation:

CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂

In order to find out if the law of conservative mass is followed, we need to count how many atoms of each element are there in both sides of the equation:

• Ca ⇒ 1 on the left, 1 on the right.

• C ⇒ 1 on the left, 1 on the right.

• O ⇒ 3 on the left, 3 on the right.

• H ⇒ 2 on the left, 2 on the right.

• Cl ⇒ 2 on the left, 2 on the right.

As the numbers for all elements involved are the same, the law is observed in the given equation.

Answer: The concentration of is 0.0122 M.

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is

We are given:

Putting values in above equation, we get:

Hence, the concentration of is 0.0122 M.