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2 years ago

How does the vapor pressure of water at 10 c compare with its vapor pressure at 50 c?

1 answer:

4 0

The vapor pressure of water at 50ºC will be greater than that at 10ºC because of the added energy and thus greater movement of the water molecules. If one knows the ∆Hvap at a given temperature, one can calculate the vapor pressure at another temperature. This uses the Clausius-Clapeyron (sp?) equation. It turns out the vapor pressure of water at 10º is 9.2 mm Hg, and that at 50º is 92.5 mm Hg.

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Explain the role of each factor in chemical weathering (water, acid, and air).

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Answer: hydrolysis oxidation acidification

Explanation:

they break down, dissolve, or create new elements

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1 year ago

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What are the two main factors that affect deep ocean currents

Alex_Xolod [135]

Answer:

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2 years ago

The zinc within a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (s

Mrac [35]

Answer:

Mass of hydrogen gas evolved is 0.0749 grams.

Explanation:

Total pressure of the gases = p = 758 mmHg

Vapor pressure of water = 23.78 mmHg

Pressure of hydrogen gas ,P = p - 23.78 mmHg = 758 mmHg - 23.78 mmHg

P = 734.22 mmHg = $\frac{734.22}{760} atm=0.966 atm$

Temperature of of hydrogen gas ,T= 25°C =298.15 K

Volume of hydrogen gas = V = 0.949 L

Moles of hydrogen gas =n

PV = nRT (Ideal gas equation )

$n=\frac{PV}{RT}=\frac{0.966 atm\times 0.949 L}{0.0821 atm L/mol K\times 298.15 K}$

n = 0.03745 mol

Moles of hydrogen gas = 0.03745 mol

Mass of 0.03745 moles of hydrogen gas = 0.03745 mol × 2 g/mol = 0.0749 g

Mass of hydrogen gas evolved is 0.0749 grams.

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3 years ago

Please help!! my grades dying <br> 15 points

AysviL [449]

Answers-in-bold:

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3 years ago

Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc

Anni [7]

Answer:

Ka = 4.76108

Explanation:

CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

[ ]initial change [ ]eq

CO(g) 0.27 M 0.27 - x 0.27 - x

H2(g) 0.49 M 0.49 - x 0.49 - x

CH3OH(g) 0 0 + x x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

7 0

3 years ago