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aalyn [17]
2 years ago
14

A watt of is a unit of power equivalent to 1 joule per second (1 J/sec). In other words, a 60 watt light bulb consumes 60 joules

of energy in a single second.
If a chocolate bar containing 200. Calories were burned and 100% of the energy released were used to a power a 100. watt laptop, how long could the chocolate bar power the laptop in hours?
Chemistry
1 answer:
Juli2301 [7.4K]2 years ago
7 0

The time the chocolate bar could power the laptop in hours is 0.00233 hrs.

Since 200 Calories of chocolate bar were burned to power the 100 Watt laptop, we need to find the number of joules on energy in 200 calories of chocolate bar.

Knowing that 4.2 Joules = 1 Calorie, then

200 Calories = 200 × 1 calorie = 200 × 4.2 Joules = 840 Joules

Since the power required by the laptop is 100 W = 100 J/s and Power, P = energy/time

so, time = energy/power

So, the time for the laptop to use 840 J of energy from the chocolate bar at a rate or power of 100 W = 100 J/s is

time = 840 J ÷ 100 J/s = 8.4 s

So, the time in hours is 8.4 s ÷ 3600 s/1 h = 0.00233 hrs (since 1 hr = 3600 s)

So, the time the chocolate bar could power the laptop in hours is 0.00233 hrs.

Learn more about time to power here:

brainly.com/question/17732603

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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

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A dehydration reaction starting with 3.8 g cyclohexanol produces 2.6 g cyclohexene. Calculate the theoretical yield for this rea
diamong [38]

Answer:

Theoretical yield of C6H10 = 3.2 g.

Explanation:

Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.

Equation of the reaction

C6H11OH --> C6H10 + H2O

Moles of C6H11OH:

Molar mass of C6H110H = (12*6) + (1*12) + 16

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Mass of C6H10 = 3.8 g

number of moles = mass/molar mass

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Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.

Therefore, 0.038 moles of C6H10 was produced.

Mass of C6H10 = molar mass * number of moles

Molar mass of C6H10 = (12*6) + (1*10)

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Mass = 82 * 0.038

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The mass of the atoms is concentrated in the very tiny space represented by the nucleus. Of course the electrons have a mass too, but is very small compared to the protons and neutrons, and we usually neglect its mass.

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