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bulgar [2K]
3 years ago
10

PLEASE HELP!!! I am taking a college chem class as a junior and have no idea how to do this!

Chemistry
1 answer:
kondaur [170]3 years ago
4 0
Where’s the questions hun? I don’t even see the questions you need answered
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Alex dragged a log across the yard in 30 the log weighted 400n she did 900 j of work how much power did I she have
In-s [12.5K]

The power used by Alex to drag the log across the yard is determined as 2,656 W.

<h3>Mass of the log</h3>

The mass of the log is calculated as follows;

W = mg

m = W/g

m = (400)/9.8

m = 40.82 kg

<h3>Velocity of the log</h3>

K.E = ¹/₂mv²

v² = 2K.E/m

v² = (2 x 900)/(40.82)

v² = 44.096

v = 6.64 m/s

<h3>Power used by Alex</h3>

P = Fv

P = 400 x 6.64

P = 2,656 W

Learn more about power here: brainly.com/question/13881533

#SPJ1

5 0
2 years ago
How much of a sample remains after three half-lives have occurred? 1/16 of the original sample 1/9 of the original sample 1/8 of
Drupady [299]
For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.
5 0
3 years ago
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How do you prepare copper sulphate​
Elena-2011 [213]
Prepare a 1% copper sulfate solution. To make this solution, weigh 1 gram of copper sulfate (CuSO4 ·5H2O), dissolve in a small amount of distilled water in a 100 ml volumetric flask and bring to volume. Label this as 1% copper sulfate solution.
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3 years ago
"You have a solution of glucose in water that has a concentration of 2.50 M and a volume of 0.442 liters. You dilute this soluti
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Answer:

0.737M

Explanation:

C1V1=C2V2

2.50 x 0.442 = C x 1.5

1.105 = C x 1.5

C = 1.105/1.5

C = 0.737M

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3 years ago
Which electrolyte is used in an alkali fuel cell?
zaharov [31]
 i think..
potassium hydroxide..
4 0
3 years ago
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