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aalyn [17]
3 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]3 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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g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial con
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In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

t_{1/2}=\frac{0.693}{k}} .....(1)

Rate law expression for first order kinetics is given by the equation:

t=\frac{2.303}{k}\log\frac{[A_o]}{[A]} ......(2)

where,  

k = rate constant

t_{1/2} =Half life of the reaction = 2.42\times 10^3 s

t = time taken for decay process = ?

[A_o] = initial amount of the reactant = 0.163 M

[A] = amount left after time t =  66.8% of [A_o]

[A]=\frac{66.8}{100}\times 0.163 M=0.108884 M

k=\frac{0.693}{2.42\times 10^3 s}

t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}

t = 1,409.19 s

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t=\frac{1,409.19 }{60} min=23.49 min

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

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Answer:

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Explanation:

<h2>hope it helps you </h2><h2>.</h2><h2>.</h2><h2>.</h2><h2>.</h2><h2>.</h2><h2>.</h2><h2>Thank uu</h2>
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