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Lana71 [14]
3 years ago
15

What is the pOH of a solution with [OH'] = 9,0 x 10-7?

Chemistry
1 answer:
bearhunter [10]3 years ago
6 0

Answer: D (6.04) is the best answer

Explanation:

pOH = - log [OH⁻]

pOH = - log (9.0 x 10⁻⁷)

pOH = 6.0457

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What is a zone of weak, variable winds located at 30 degrees north and 30 degrees south?
timurjin [86]

Answer:

Horse latitude, trade winds

Explanation:

  • The area of the low pressure or the calm consists of the variable light winds that blow near the equator are known to the marines as the doldrums and they form a circuital pattern near the earth atmosphere.
  • Forms at a center of the near the higher pressure systems called as the horse latitudes where the  trade winds at the surface are weak and variable and this zone is found generally in latitudes of the 30° North and South of the equator and move in an east to west direction.

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3 years ago
Help and explain step by step if you can!
Leona [35]

Answer:

the third one

Explanation:

if you look at it mathematically its the one that makes the most sense

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Norma-Jean [14]

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6 0
2 years ago
Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co
amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as p_{d} and the proportion of air that is by-passed as p_{bp}, being p_{d}+p_{bp}=1.

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

0.004 = 0.016*p_{bp}+ 0.002*p_{d}

Replacing the first equation in the second one we have

0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0
3 years ago
Aonde a civilização egípcia se desenvolvera​
Lorico [155]
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