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KonstantinChe [14]
2 years ago
12

A gas occupies 1.00 L at standard temperature. What is the volume at 606 K?

Mathematics
1 answer:
tino4ka555 [31]2 years ago
3 0

Answer:

the volume at 606k is 2.22....the third one

Step-by-step explanation:

v1=1.00 t1=273 v2=? t2=606

v1/t1=v2/t2

1.00/273=v2/606

1.00×606=273v2

606=273v2

606/273=v2

2.21 L=v2

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Pls help find the amount at the end of 6 hours
marshall27 [118]

Answer:

A. 7,348

Step-by-step explanation:

P = le^kt

intitial population = 500

time = 4 hrs

end population = 3,000

So we have all these variables and we need to solve for what the end population will be if we change the time to 6 hours. First, we need to find the rate of the growth(k) so we can plug it back in. The given formula shows a exponencial growth formula. (A = Pe^rt) A is end amount, P is start amount, e is a constant that you can probably find on your graphing calculator, r is the rate, and t is time.

A = Pe^rt

3,000 = 500e^r4

now we can solve for r

divide both sides by 500

6 = e^r4

now because the variable is in the exponent, we have to use a log

log_{e}(6) = 4r

ln(6) = 4r

we can plug the log into a calculator to get

1.79 = 4r

divide both sides by 4

r = .448

now lets plug it back in

A = 500e^(.448)(6 hrs)

A = 7351.12

This is closest to answer A. 7,348

4 0
2 years ago
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Lostsunrise [7]

Answer:

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Step-by-step explanation:

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k =  \frac{2}{1}

z =  \frac{1}{2} q

z =  \frac{q}{2}

8 0
3 years ago
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maksim [4K]

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Step-by-step explanation:

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I believe C. 5/9 is correct
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3 years ago
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