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Alisiya [41]
3 years ago
8

Many times, during a chemical reaction, energy is given off. this type of chemical reaction can be termed

Chemistry
1 answer:
WITCHER [35]3 years ago
3 0
Physical can be seen
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To fill the other element's shell.
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1. What is the Charlotte?<br> A. A train<br> B. A car<br> D. An airplane<br> C. A ship
dybincka [34]

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B is the Charlotte

Explanation:

it is Charlotte because it has 4 wheels

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Which of the following is NOT true about wastewater and wastewater treatment?
Mars2501 [29]

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4 0
3 years ago
Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25
o-na [289]
The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell 
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V 
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation. 
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72
6 0
3 years ago
Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d
worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.  

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B\rightarrow Products

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

3.20\times 10^{-1}=k(1.50)^a(1.50)^b ....(1)

Expression for rate law for second observation:

3.20\times 10^{-1}=k(1.50)^a(2.50)^b ....(2)

Expression for rate law for third observation:

6.40\times 10^{-1}=k(3.00)^a(1.50)^b ....(3)

Dividing 1 from 2, we get:

\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0

Dividing 1 from 3, we get:

\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[A]^1[B]^0

\text{Rate}=k[A]

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0
3 years ago
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