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1 year ago

14.28 explain how you would distinguish between each pair of compounds using high- resolution mass spectrometry

1 answer:

7 0

The distinguish between each pair of compounds using high- resolution mass spectrometry by the exact mass rather than nominal mass are utilizes to measure the compound.

The mass spectrometry is involves the following steps :

The ionization
acceleration
deflection
detection

Mass spectrometry is the analytical method useful for the calculating the mass to charge ratio ( m / z ). the mass spectrometry is based on the newton's second law and the momentum.

Thus, the mass spectroscopy is method to measure the molecular mass of the compound and indirectly helps examine the isotopes and based on the newton's second law .

To learn more about mass spectroscopy here

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Arrange these elements in order of decreasing atomic size:sulfur, chlorine, aluminium, and sodium.

Alborosie

Answer:

Sodium,aluminium,sulfur and chlorie

7 0

3 years ago

) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7

Travka [436]

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

$2O_3\rightleftharpoons 3O_2$

Rate of formation of oxygen : $7.78\times 10^{-1} M/s$

Rate of the reaction(R) = $\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}$

$R=\frac{1}{3}\frac{d[O_2]}{dt}$

Rate of formation of oxygen=3 × (R)

$7.78\times 10^{-1} M/s=3\times R$

Rate of the reaction(R): $0.2593 M/s$

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

$R=-\frac{1}{2}\frac{d[O_3]}{dt}$

$\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s$

-0.5186 M/s is the rate of the loss of ozone.

6 0

3 years ago

N76 [4]

Answer:

Explanation:

4 0

3 years ago

or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und

beks73 [17]

Answer:

The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

$C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)$

$\Delta H^{o}_{rxn}= -5104.1kJ/mol$

The expression for the entropy change for the reaction is as follows.

$\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]$

$\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol$

$\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol$

$\Delta H^{o}_{f}(O_{2})= 0kJ/mol$

Substitute the all values in the entropy change expression.

$-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]$

$-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})$

$\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol$

$=-220.1kJ/mol$

Therefore, The standard enthalpy of formation of this isomer of $C_{8}H_{18}$ is -220.1 kJ/mol.

4 0

3 years ago

Calculate the concentration of all species in a 0.160 m solution of h2co3.

enot [183]

H₂CO₃ ⇔ HCO₃⁻ + H⁺

I 0.160 0 0

C -x +x +x

E 0.160-x +x +x

Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]

4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)

x² = 6.88 x 10⁻⁸

x = 2.62 x 10⁻⁴

HCO₃⁻ ⇔ CO₃⁻² + H⁺

I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴

C -x +x +x

E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x

Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]

5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)

x = 5.6 x 10⁻¹¹

Thus,

[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M

[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M

[CO₃⁻²] = 5.6 x 10⁻¹¹ M

[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M

[OH⁻] = 3.8 x 10⁻¹¹

8 0

3 years ago