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Leya [2.2K]
3 years ago
11

What is the dominant greenhouse component and why?

Chemistry
1 answer:
insens350 [35]3 years ago
7 0

Answer: HOPE THIS HELPED! :D

Water vapor is the most abundant and dominant GHG in the atmosphere. Its concentration depends on temperature and other meteorological conditions, and not directly upon human activities.

Explanation:

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Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la
tatuchka [14]

Answer:

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]

2. Derive the rate law

Assume k₋₁ ≫ k₂.  

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.  

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is  

rate = k[A][B] where k = k₂K

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What is Avagrados Number?
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What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of
weqwewe [10]

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{31.862\ g}{63.546\ g/mol}

<u>Moles of copper = 0.5014 moles </u>

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,  

Mass of nitric acid = \frac {68}{100}\times 284\ g = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{193.12\ g}{63.01\ g/mol}

<u>Moles of nitric acid = 3.0649 moles </u>

According to the reaction,  

Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}

1 mole of copper react with 4 moles of nitric acid

Thus,  

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

<u>Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

<u>Moles of nitrogen dioxide = 1.0028 moles </u>

Given:  

Pressure = 735 torr

The conversion of P(torr) to P(atm) is shown below:

P(torr)=\frac {1}{760}\times P(atm)

So,  

Pressure = 735 / 760 atm = 0.9632 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.2 + 273.15) K = 301.35 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K  

<u>⇒V = 25.76 L</u>

4 0
3 years ago
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