All categories

4 years ago

Three facts about the troposphere

Chemistry

2 answers:

joja [24]4 years ago

7 0

Sorry all i know is one and it is (<span>Contains almost all of the atmospheric water vapor)</span>

madam [21]4 years ago

4 0

1) Troposphere is the nearest layer to the earth.

2) Range from 5 to 11 miles in thickness.

3) Contains almost all of the atmospheric water vapor.

You might be interested in

All methods of chromatography operate on the same basic principle that Select one: a. one component of the mixture will chemical

ANEK [815]

Answer:C

Explanation:

Chromatography separates compounds by taking advantage of their polarity. The stationary phase is generally very polar. The mobile phase can be pure hexane or various ratios of hexane with a polar eluent added. The more polar the compound, the more it interacts with the stationary phase and won’t move very far up the plate compared to the non-polar or less polar compounds that interact more with the non-polar hexane.

8 0

3 years ago

Can someone good with chemistry help me?

Digiron [165]

As you go down group 7 the melting point of the elements will increase, this is because as you go down the group you are gaining an electron shell and the molecule will become bigger. This increase in size means that there will be an increase in the intermolecular forces as well.

So the order would be NF3, NBr3, NI3.

Hopefully this helps!

6 0

3 years ago

1 point<br> Convert 3.79 x 10^24 atoms of sodium to grams.

bija089 [108]

Answer:

is easy to poiling

Explanation:

also is dificult to melting

3 0

2 years ago

What color does red cabbage juice make when mixed with a citrus cleaner?

Nina [5.8K]

Answer: PINK

Explanation:

7 0

2 years ago

Read 2 more answers

0.0500 mol of gas occupies a cylinder which is sealed on top by a moveable piston. The piston is circular, with a mass of 30.0 k

Viefleur [7K]

Answer:

The workdone by both N₂ and neon gas is 49.3 J

The change in internal energy of N₂ and neon gas is 125.6 J and 73.54 J respectively

The heat for N₂ and neon gas is 171.9 J and 122.84 J respectively.

Explanation:

Given that:

number of moles = 0.05 mole

mass of the piston = 30 kg

diameter = 5.00 cm = 0.05 m

Area (A) = πr²

$Area (A) = \pi*(\frac{0.05}{2})^2 \\ \\ Area (A) = 0.0019635 \\ \\ Area (A) = 19.635*10^{-4} \ m^2$

The piston is said to move from 30 cm - 40 cm

So, the change in volume ΔV is calculated as:

$=(40-30)*10^{-2} *19.635*10^{-4}$

$= 1.9635*10^{-4} \ m^3$

Outside the cylinder; the pressure $P_{air}= 1 \ atm$ = 101325 Pa

Thus, workdone $w_1$ = PΔV

= $101325*1.9635*10^{-4}$

= 19.90 J

The gravitational work $w_2 = mgh$

Given that the height (h) = 10 cm = 0.1 m

Then; $w_2 = 30*9.8*0.1$

$w_2 = 29.4 \ J$

The total workdone $w_{total}$ for both cases is:

$w_{total } =w_1 + w_2$

$w = (19.90 + 29.4) \ J$

$w =49.3 \ J$

The pressure of gas inside the cylinder is determined as:

$P_{in}.A = P_{out}.A +mg$

$(P_{in}-P_{out}) = \frac{mg}{A} \\ \\ P_{in} -10^5 = \frac{30*9.8}{19.635*10^{-4}} \\ \\ P_{in} = 149732.6203+10^5 \\ \\ P_{in} = 2.497*10^5 \ Pa$

a). assuming that the gas is N₂.

$C_v =\frac{5}{2}R$

Thus, the change in internal energy ΔU is given as:

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{5}{2}R \delta T \\ \\ \delta U = \frac{5}{2}nR \delta T$

Since $P_{in} \delta V = nR \delta T ; \ Then$ ;

$\delta \ U = \frac{5}{2} P_{in} \delta V \\ \\ \delta \ U = \frac{5}{2}*2.497*10^5 *1.9635*10^{-4} \\ \\ \delta \ U = 122.57 \ J$

ΔU ≅ 125.6 J

The heat Q = ΔU + W

Q = (122.6 + 49.3) J

Q = 171.9 J

b) In Neon gas:

$C_v = \frac{3}{2}R$

∴

change in internal energy is;

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{3}{2}R \delta T \\ \\ \delta U = \frac{3}{2}P_{in}.V$

$\delta U = \frac{3}{2}*2.497*10^5*1.9635*10^{-4}$

ΔU = 73.54 J

The heat Q = ΔU + W

Q = (73.54 + 49.3) J

Q = 122.84 J

5 0

3 years ago