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2 years ago

Describe how you might determine the m/z and relative abundance of the ions contributing to the peak at 21.876 min

1 answer:

8 0

The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.

<h3>What is relative abundance?</h3>

The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.

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1.00 mole of an ideal monoatomic gas at STP first undergoes isothermal expansion so that the volume at b is 2.5 times the volume

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the pressure at c = 0.27 atm

Explanation:

Given that:

number of moles (n) = 1.0 moles

Value of gamma in the monoatomic gas (γ) = 5/3

During an isothermal expansion, the volume at b is = 2.5 times the volume at a ; this implies that:

$V_b = 2.5 V_a$

∴ To calculate the pressure at c from a; the process is adiabatic compression; so we apply:

$P_aV_a^\gamma=P_cV_c^\gamma$

$\frac{P_c}{P_a}=[\frac{V_a}{V_c}]^{(2/3)$

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3 years ago

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

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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

5 0

4 years ago