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ddd [48]
3 years ago
10

How does the latent heat of evaporation (and its equivalent, the latent heat of condensation) moderate climate?

Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

By absorbing energy on evaporation and releasing it on condensation, it keeps water cool when the air is hot and warm when the air is cool.

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The electronic configuration of phosphorous is 2.8.5. Explain, in terms of its electronic configuration, why phosphorus is in gr
Setler [38]

Answer:

Explanation:

phosphorus belongs to group 5 of the periodic table because it has 5 electron in its outermost shell the number of electron in the outermost shell of electron determine the group of the element in the periodic table

6 0
3 years ago
C₇H₆O₂ + O₂ -->CO₂ +H₂O Find the chemical reaction
lions [1.4K]

Answer:

Air

Explanation:

idrk im just an idiot

3 0
2 years ago
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1.For the reaction P4 O10(s) + 6H2O(l) → 4H3PO4(aq), what mass of P
Alex17521 [72]
P₄O₁₀ + 6H₂O → 4H₃PO₄
The equation shows us that the molar ratio of
P₄O₁₀ : 6H₂O = 1:6

We also know that one mole of a substance contains 6.02 x 10²³ particles. We can use this to calculate the moles of water.
moles(H₂O) = (5.51 x 10²³) / (6.02 x 10²³)
= 0.92 mole
That means moles of P₄O₁₀ = 0.92 / 6
= 0.15

Each mole of P₄O₁₀ contains 4 moles of P. 
moles(P) = 4 x 0.15 = 0.6 mol
Mr of P = 207 grams per mol
Mass of P = 207 x 0.6
= 124.2 grams
5 0
3 years ago
Read 2 more answers
Simplify: <br>(100 m)/(26 s)
podryga [215]

100m/26s=50m/13s

50m/13s=3.846m/s

4 0
3 years ago
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Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)
Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

S°(H₂O) = 69.91 J/K.mol

S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0
3 years ago
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