A light bulb consumes energy at a rate of 80 joules per second . How long in

seconds will it take for the light bulb to consume 2.30 ~ 105

in energy?

2875 s

Explanation:

From the question given above, the following data were obtained:

Power (P) = 80 J/s

Energy (E) = 2.3×10⁵ J

Time (t) = ?

Power can simply be defined as the rate at which energy is used. Mathematically, power is expressed as:

Power = Energy / time

P = E/t

With the above formula, we can obtain the time taken for the bulb to consume 2.3×10⁵ J. This can be obtained as follow:

Power (P) = 80 J/s

Energy (E) = 2.3×10⁵ J

Time (t) = ?

Power = Energy / time

P = E/t

80 = 2.3×10⁵ /t

Cross multiply

80 × t = 2.3×10⁵

Divide both side by 80

t = 2.3×10⁵ / 80

t = 2875 s

Therefore, it will take 2875 s for the bulb to consume 2.3×10⁵ J of energy.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.