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konstantin123 [22]
3 years ago
11

Three of the primary components of air are

Chemistry
1 answer:
Darya [45]3 years ago
8 0

Answer:

125.681 torr

Explanation:

The formula to calculate the partial pressure of oxygen:

P(O2) = Pcommon - PCO2 - PN2

1 atm = 760 torr

Therefore you need to find the partial pressure of oxygen:

P(O2) = 760 torr -0.285 torr - 634.034 torr =130.013 torr

- Hope that helped!

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Why would it be better to be an r-selected species if the water resources in an area were to become more limited over a short
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R selected species are better equipped to survive environmental change because their quick population turnover can quickly adapt.

Explanation:

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A piston compresses the air in the cylinder by doing 67.0 J of work on the air. As a result, the air gives off 149.0 J of heat t
harina [27]

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Which of the following compounds contains the tin (IV) ion? A. SnO2 B. SnCl2 C. Sn2O D. Sn4O3
Zolol [24]
<span>A. SnO2 is the correct answer. I hope this helps </span>
3 0
3 years ago
Read 2 more answers
How many atoms of lead are in 3.25 moles of lead
musickatia [10]

Answer: 1.96x10^24 atoms

Explanation:

3.25*6.02214076*10^23 atoms = 1.96x10^24

6 0
3 years ago
A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt
Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
6 0
3 years ago
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