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Crazy boy [7]
3 years ago
9

Which of the following is not part of our climate system?

Chemistry
1 answer:
Afina-wow [57]3 years ago
4 0
I think the ones that don’t belongs is Cities and ocean
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How many liters are present in 10 grams of H2O ?
Luba_88 [7]

Answer:

About 0.01

Explanation:

1 grams of H2O = 0.0010

0.0010 x 10 = 0.01

8 0
3 years ago
If 4520 kj of heat is needed to boil a sample of water, what is the mass of water
Cerrena [4.2K]

Answer:

1,085g of water

Explanation:

If we have the value 4520kj is because the question is related to Energy and heat capacity. In this case, the law and equation that we use is the following:

                                                  Q= m*C*Δt  where;

Q in the heat, in this case: 4520kj

m is the mas

Δt= is the difference between final-initial temperature (change of temperature), in this exercise we don´t have temperatura change.

In order to determine the mass, I will have the same equation but finding m

                                          m= Q/C*Δt    without   m=Q/C

So: m= 4,520J/4.18J/g°C

      m= 1,0813 g

5 0
3 years ago
How many moles of ions would you expect in an aqueous solution containing one mole of chromium(III) chloride? Hint: write out th
Cerrena [4.2K]

Answer:

Four  

Explanation:

AlCl₃(aq) ⟶ Al³⁺(aq) + 3Cl⁻(aq)

One mole of AlCl₃  produces 1 mol of Al³⁺ and 3 mol of Cl⁻.

That's four moles of ions.

8 0
3 years ago
(Thermodynamics)
frutty [35]

Answer:

3853 g

Step-by-step explanation:

M_r: 107.87

         16Ag + S₈ ⟶ 8Ag₂S; ΔH°f =  -31.8 kJ·mol⁻¹

1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag

You must react 3853 g of Ag to produce 567.9 kJ of heat

3 0
3 years ago
4. What type of evidence do you think would be most difficult to collect
mars1129 [50]

I think it is trace evidence since it is really small and hard to find.

3 0
4 years ago
Read 2 more answers
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