To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows
time 0 40 80 120 160
moles 0.100 0.067 0.045 0.030 0.020
Q1)
for the first 40 s change of moles ;
= -d[A] / t
= - (0.067-0.100)/40s
= 8.25 x 10⁻⁴ mol/s
for the next 40 s
= -(0.045-0.067)/40
= 5.5 x 10⁻⁴ mol/s
the 40 s after that
= -(0.030-0.045)/40 s
= 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction
rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction
Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation.
Since this is a first order reaction,
b = 1
therefore the reaction is
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
= 8.25 x 10⁻³ s⁻¹
An atom gains an electron from another atom. Hence, option B is the correct answer.
<h3>What is an atom?</h3>
An atom is a particle of matter that uniquely defines a chemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons.
When an atom shares electrons with another atom then it results in the formation of a covalent bond.
Whereas when an atom transfer electrons from one atom to another then it results in the formation of an ionic bond.
When the nucleus of an atom splits then it represents a nuclear fission reaction and energy is released during this process.
Hence, option B is the correct answer.
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The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
Li⁰(s) → Li⁺(aq) + e⁻ (2)
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
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Answer:
the concentration of the solution is 0.00906 M
Explanation:
Given the data in the question;
we know that from Nernst Equation;
E = E⁰ - ((0.0592/n) logQ)
now, E₀ for concentration cell is 0
n for this redox is 2
concentration of the unknown solution is x
so we substitute
0.045 = 0 - ( 0.0592 / 2)log( x/0.300 ))
0.045 = -0.0296log( x/0.300 )
divide both side by 0.0296
1.52 = -log( x/0.300 )
x/0.300 =
x/0.300 = 0.0301995
we cross multiply
x = 0.300 × 0.0301995
x = 0.00906 M
Therefore, the concentration of the solution is 0.00906 M
Answer:
The right answer is "
".
Explanation:
The given values are:

According to the question,
The concentration of standard will be:
= 
= 
= 
Coffee after dilution,
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
hence,
In unknown sample, the concentration of coffee will be:
= 
= 
= 