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Scorpion4ik [409]

2 years ago

10

the water in a tea pot is heated on a stove top. The temperature of the water increases. Is this an endothermic or exothermic pr

ocess?

1 answer:

marusya05 [52]2 years ago

5 0

It is an endothermic process

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How many times thus the earth fit inside the sun

vodka [1.7K]

More than one million Earths could fit inside the Sun if it were hollow. The Sun has a radius of 696.340 km / 432.685 mi and a diameter of 1.39 million km / 864.000 mi.

6 0

2 years ago

c) A substance has a high melting point and conducts electricity. What type of structure does it have

trapecia [35]

Answer:

Metallic structure

Explanation:

They have a high melting point due to the strong forces of attraction between the positive ions (cations) and the delocalised electrons. Moreover, they conduct electricity due to the sea of delocalised electrons.

<em>[Extra: It could be an ionic compound since they also have a high melting point, however they only conduct electricity in liquid or aqeouus state.]</em>

6 0

1 year ago

CAN SOMEONE PLEASE HELP ME?!

Len [333]

Answer:

no

Explanation:

because

5 0

2 years ago

2 icl + h2i2 + 2 hcl is first order in icl and second order in h2. Complete the rate law for this reaction in the box below. Use

OleMash [197]

2 ICl + H2 ----> I2 + 2 HCl

as given that rate is first order with respect to ICl and second order with respect to H2

The rate law will be

Rate = K [ICl] [ H2]^2

b) Given that K = 2.01 M^-2 s^-1

Concentrations are

[ICl] = 0.273 m and [H2] = 0.217 m

Therefore rate = 2.01 X (0.273)(0.217)^2 = 0.0258 M / s

5 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