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Anestetic [448]
3 years ago
9

How do greenhouse gases affect the heat flow into and out of Earth's atmosphere?

Chemistry
1 answer:
guapka [62]3 years ago
3 0

they allow solar energy to penerate the atmosphere and get absorbdd by thr earths surface thus warming it up. they absorb thr out going earths energy keeping thr energy warmth near the earths surface
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A 2.00 kg piece of lead at 40.0°C is placed in a very large quantity of water at 10.0°C,and thermal equilibrium is eventually re
sveticcg [70]

Answer:

Δ S = 26.2 J/K

Explanation:

The change in entropy can be calculated from the formula  -

Δ S = m Cp ln ( T₂ / T₁ )

Where ,

Δ S = change in entropy

m = mass  = 2.00 kg

Cp =specific heat of lead is 130 J / (kg ∙ K) .

T₂ = final temperature  10.0°C + 273 = 283 K

T₁ = initial temperature ,  40.0°C + 273 = 313 K

Applying the above formula ,

The change in entropy is calculated as ,

ΔS = m Cp ln ( T₂ / T₁ )  = (2.00 )( 130 ) ln( 283 K / 313 K )

ΔS = 26.2 J/K

6 0
3 years ago
Number 2 I don't get what the question is asking
Rus_ich [418]
Wow thats pretty hard its asking how many is in each of those
7 0
3 years ago
What volume of 0.194 MNa3PO4 solution is necessary to completely react with 85.5 mL of 0.109 MCuCl2 ?
ira [324]

Answer:

35.9 ml

Explanation:

Start with the balanced equation:

3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)

This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-

∴  1 mole CuCl2 will react with 2/3 moles Na3PO4

We know that concentration = moles/volume i.e:

c= n/v

∴n=c×v

∴nCuCl2=0.107×91.01000=9.737×10−3

I divided by 1000 to convert ml to L

∴nNa3PO4=9.737×10−3×23=6.491×10−3

v=nc=6.491×10−30.181=35.86×10−3L

∴v=35.86ml

4 0
2 years ago
Hi, May I have Chem help please? Please Keep the answer as simple as possible.
lesya [120]

Explanation:

a positively charged nucleus is surrounded by mostly empty space.

4 0
3 years ago
Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these are decomposed, the sulfur dioxide produce
Diano4ka-milaya [45]
To determine the mass of oxygen per gram of sulfur for sulfur dioxide, we simply obtain the ratio of the mass of oxygen and the mass of sulfur produced from the decomposition of sulfur dioxide. All other values given in the problem statement above are just to confuse us that the question is a difficult one. We do as follows:

mass of oxygen per gram sulfur = 3.45 g / 3.46 g
mass of oxygen per gram sulfur = 0.9971 g O2 / g S
4 0
3 years ago
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