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4 years ago

The gas absorbed by ocean water which can moderate a potential greenhouse effect is:

Chemistry

2 answers:

Eddi Din [679]4 years ago

6 0

<h3><u>Answer;</u></h3>

Carbon dioxide

The gas absorbed by ocean water which can moderate a potential greenhouse effect is <u>carbon dioxide</u>.

<h3><u>Explanation;</u></h3>

Greenhouse effect is is a process that warms the Earth's surface. This is as a result of radiation and absorption of sun's energy by greenhouse gases.

The ocean decrease the rate of accumulation of carbon dioxide in the atmosphere by absorbing carbon dioxide gas which is a green house gas.
Oceans absorbs carbon dioxide from the atmosphere where the air meets the water. The turbulence and waves caused by wind makes water to absorb the carbon dioxide.

trapecia [35]4 years ago

3 0

The gas absorbed by ocean water which can moderate a potential greenhouse effect is <span>CO2 (Carbon Dioxide). </span>

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A water tank measures 24in.×48in.×12in. Find the capacity of the water tank in cubic feet. Do not include units in your answer.

LenKa [72]

<em>Answer:</em> 8 (feet)

<em>Explanation:</em>

24 inches = 2 feet

48 inches = 4 feet

12 inches = 1 foot

To find volume you do Base * Width * Height

2*4*1 = 8

Hope this helps!

6 0

3 years ago

You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

Gala2k [10]

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound Liquid molar volume (cm³/mol)

Ethanol (1) 58.68

1 - propanol (2) 75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = $\frac{V_2}{V_1} \ exp \ (\frac{-a_{12}/R}{T} )$ ------------ equation (1)

where:

$a_{12}/R$ = Wilson parameter = ???

$V_2$ = liquid molar volume of component 2 = 75.14 cm³/mol

$V_1$ = liquid molar volume of component 1 = 58.68 cm³/mol

T = temperature = 25° C = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

$0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$In (0.547) = \ (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R= 0.60 * 298.15 \ K$

$-a_{12}/R= - 178.89 \ K$

$a_{12}/R= 178.89 \ K$

To calculate the temperature dependent parameters of the Wilson equation ∧₂₁

∧₂₁ = $\frac{V_1}{V_2} \ exp \ (\frac{-a_{12}/R}{T} )$ ---------- equation (2)

$1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1.1 = 0.7809 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$\frac{1.1}{0.7809}= exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R = 0.3426 * 298.15 \ K$

$-a_{12}/R =102.15 \ K$

$a_{12}/R = -102.15 \ K$

From equation (1) ; let replace 178.98 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = $\frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )$

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = $\frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )$

∧₂₁ = 0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

8 0

3 years ago

A forklift applies a force of 2,000 N to raise a box 3 m.

anygoal [31]

If the force is applied to lift the box, then there is no work done as the displacement of the box is not in the same direction of the box.

However, if the force is being applied to move the box in the direction of the force, then the work done is:

Work = force x distance

Work = 1500 x 3

Work = 4,500 Joules

6 0

3 years ago

Select the correct answer.

Alex

no:A

is a correct answer

7 0

3 years ago

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2 HCN + _ CuSO4 + H2SO4 + Cu(CN)2<br><br> Reaction type?

HACTEHA [7]

This is a double replacement or double displacement reaction

3 0

3 years ago