(35 kg)*(1000 g / 1 kg)
(35 kg / 1 kg)*(1000 g)
kg units cancel, you're left with units of g
35/1*1000 = 35,000
ANSWER:
35,000 g
Answer:
The cost of electricity for 100 W power bulb = $ 32.85
Cost of electricity for 0.025 W fluorescent bulb = $ 8.2125
Explanation:
Cost of electricity = $ 0.18 per KW-H
Time = 5 hour per day
Bulb power = 100 W = 0.1 KW
Fluorescent bulb power = 25 W = 0.025 KW
(a) Cost of electricity for 100 W power bulb
0.1 × 5 × 365 × 0.18 = $ 32.85
(b) Cost of electricity for 0.025 W fluorescent bulb
0.025 × 5 × 365 × 0.18 = $ 8.2125
Therefore the cost of electricity for 100 W power bulb = $ 32.85
Cost of electricity for 0.025 W fluorescent bulb = $ 8.2125
Answer:
A feasible error could have been the removal of the sample before all water evaporated.
Explanation:
In order to determine the percentage of water in an hydrate, an experiment that could be performed is the heating of the sample until the mass does not change. If the student heated the sample an insufficient amount of time, water will be present in the sample, thus reducing the percentage reported.
The answer to the question is D.
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