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3 years ago

In which of the following states of matter are the particles moving the fastest? (4 points)

Chemistry

1 answer:

xenn [34]3 years ago

7 0

Hello.

The answer is Gas.
In gases the partcails have more space and can move faster.

Have a nice day

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Hiiii everyone my previous account has been deleted who knows me plz follow

JulsSmile [24]

Answer:

okie!

Explanation:

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2 years ago

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8.5ml of a sample of sea water solution was added to a 44.317g evaporating dish the combination weighted 52.987g after evaporati

Ratling [72]

Answer: There is no question, but we can calculate a couple of items:

Density of sea water sample = (52.987g-44.317g)/8.5ml

Inorganic content of sample (mostly salts) = (44.599g-44.317g)/(52.987g-44.317g) x 100% = percent inorganics in water sample

Explanation:

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2 years ago

The decomposition of nitryl chloride is described by the following chemical equation: 2NO2C1(g) → 2NO2 (g)-C12 (g) Suppose a two

Gnoma [55]

Answer:

Second reaction

NO2 + F -------> NO2F

Rate of reaction:

k1 [NO2] [F2]

Explanation:

NO2 + F2 -----> NO2F + F slow step1

NO2 + F -------> NO2F fast. Step 2

Since the first step is the slowest step, it is the rate determining step of the reaction

Hence:

rate = k1 [NO2] [F2]

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3 years ago

Which of the following is NOT an effect of human development on the Everglades?

Fofino [41]

Answer: I think its A

Explanation:

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3 years ago

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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