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3 years ago

What is the overall reaction order for this rate law: rate = k[A]2[B][C]?

Chemistry

2 answers:

dybincka [34]3 years ago

8 0

Zero order are reactions in which concentration of reactant has NO effect on RATE OF REACTION.

2. First order are reactions in which concentration of one reactant is proportionate to the RATE OF REACTION.

Exp: That means when you increase the concentration of the one reactant, then the rate of reaction will increase by the same degree of extent.

3. Second order are reactions in which concentration of two reactant has an effect on the RATE OF REACTION.

Formula:

1. Zero order Rate = k

2. First order Rate = k(A)^m

3. Second order Rate = k(A)^m(B)^n

where () represents concentration

and equation is mA + nB -> Product.

KonstantinChe [14]3 years ago

5 0

Answer:

The overall reaction order for this rate law is 4.

Explanation:

The order of the reaction is the sum of power of the concentrations of reactants.

$rate=k[x]^a[y]^b$

The order of the reaction = a+b

According to reaction , the rate of the reaction is:

Rate = k[A]2[B][C]

Order of the reaction = 2 + 1 + 1= 4

The overall reaction order for this rate law is 4.

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Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6

Sergeeva-Olga [200]

Explanation:

The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

= 4.11 M

Hence, molarity of toulene is 4.11 M.

As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

= 8.6 m

Hence, molality of given toulene solution is 8.6 m.

Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{75.8 g}{92.13 g/mol}$

= 0.8227 mol

Now, no. of moles of benzene will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{95.6 g}{78.11 g/mol}$

= 1.2239 mol

Hence, the mole fraction of toulene is as follows.

Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

= $\frac{0.8227 mol}{(0.8227 + 1.2239) mol}$

= 0.402

Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

0.857 $g/cm^{3}$ = $\frac{mass}{200 ml}$ (As 1 $cm^{3}$ = 1 g)

mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

= 44.22%

Therefore, mass percent of toulene is 44.22%.

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3 years ago

Помогите пожалуйста или с первым вопросом или со вторым. Буду очень благодарна, т.к. это мой зачет... :С

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Answer: D

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3 years ago

The gas in an engine cylinder is compressed until it has a volume of 0.045 liter and an absolute pressure of 28 atmospheres. If

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$7.7 \times 10^{2} \; \text{K}$ .

<h3>Explanation</h3>

Units of each quantity:

<em>P</em> = 28 atm;
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<em>n</em> = 0.020 mol;

Ideal gas constant:

$R = 0.0820573 \;\text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$ .

Apply the ideal gas law:

$T = \dfrac{P \cdot V}{n \cdot R} = \dfrac{{\bf 28} \; \text{atm} \times {\bf 0.045} \;\text{L} }{{\bf 0.020}\; \text{mol} \times {\bf 0.0820573} \;\text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}} = {\bf 767.76} \; \text{K}$ .

All data given in this question come with two significant figures. Round the value of <em>T</em> to two significant figures:

$T = 767.76 \;\text{K} = 7.7 \times 10^{2} \; \text{K}$ .

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