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nevsk [136]
3 years ago
6

Now let’s return to the question of why everything went off when the light, a heater, and a hair dryer were all running in the b

athroom at the same time. After all, the light usually stays on just fine whenever you use the bathroom. Why would it go off now?
Answer this question assuming the circuit breaker or fuse for the bathroom has a 40-ampere capacity. Use the terms “circuit,” and “overload.” Then think of at least two solutions to prevent this from happening again.
Chemistry
2 answers:
igomit [66]3 years ago
6 0

Answer:

The power must have gone out for that area.

Explanation:

erastova [34]3 years ago
3 0

Answer:

the amount of power went over the circuit breaker limit

Explanation:

The circuit breaker cut off everything in the bathroom because the limit was 40 amperes but the amount of power being used in the bathroom exceeded the limit, going up to 46.25 amperes. A way to stop this may be using an extension cord to plug in the hair dryer somewhere outside the bathroom while still being able to use it in the room, or instead of using the light in the bathroom, try using a night light or just light from outside the room that can still provide enough light in the bathroom.

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4 0
3 years ago
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A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

3 0
3 years ago
How many moles of CaCl2 are contained in 0.448 L of a 0.85 M CaCl2 solution?
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3 0
3 years ago
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3 years ago
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