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Svetllana [295]
2 years ago
5

What is the rate constant of a reaction if rate = 1 x 10-2 (mol/L)/s, [A] is 2 M,

Chemistry
2 answers:
boyakko [2]2 years ago
8 0

The rate constant of a reaction : 8.3 x 10⁻⁴

<h3>Further explanation</h3>

Given

rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M,  [B] is 3 M, m = 2, and n = 1

Required

the rate constant

Solution

For aA + bB ⇒ C + D

Reaction rate can be formulated:

\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}

the rate constant : k =

\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}

Firlakuza [10]2 years ago
8 0

Answer:

8.3x10^10-4

Explanation:

a p e x

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Electrical energy is the energy of__________
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Answer:

<h3>electrical energy is the energy of Kinetic energy </h3>

Explanation:

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6 0
2 years ago
Read 2 more answers
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

Now put all the given values in the above formula, we get:

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

Now put all the given values in the above formula, we get:

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

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3 years ago
Quiz
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4 0
3 years ago
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If 4.0 mol of NO and 4.0 mol of O2 are combined, how many moles
Masja [62]
4.0


i think it has something to do with molar ratios and finding the limiting reactant

4.0 mol NO * 2 mol NO2/2 mol NO = 4.0 moles of NO2

4.0 mol O2 * 2 mol NO2/1 mol O2 = 8.0 moles of NO2

so the limiting reactant (the reactant that runs out the quickest leaving an excess) is NO

once the limiting reactant is found, we can use that data for that substance to calculate the amount of product

4.0 mol NO * 2 mol NO2/2 mole NO = 4.0 moles of NO2

4 0
3 years ago
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