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7 days ago

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Consider the ammonolysis of benzoyl chloride by adding concentrated ammonium hydroxide to form the final product, benzamide. Rea

ction scheme of benzoyl chloride with cold, conc. ammonium hydroxide over ther reaction arrow, and benzamide, hydrochloric acid, and water as products. Calculate the molar mass of the reactant and product. Report molar masses to 1 decimal place. Molar mass of benzoyl chloride g / mol Molar mass of benzamide g / mol

1 answer:

Alekssandra [2.7K]7 days ago

4 0

Answer:

Explanation: refer to the attached document below

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Write a balanced equation for the reaction of NaCH3COO (also written as NaC2H3O2) and HCl.

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The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan

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Response: The rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

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Based on the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

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$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

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$K_1$ = rate constant when $701K$ = $2.57M^{-1}s^{-1}$

$K_2$ = rate constant when $525K$ =?

$Ea$ = activation energy for the process = $1.5\times 10^2kJ/mol=1.5\times 10^5J/mol$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 701 K

$T_2$ = final temperature = 525 K

Substituting the provided values into this formula yields:

$\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]$

$K_2=0.0606M^{-1}s^{-1}$

Thus, the rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

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20 days ago

Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass and has a molar mass of 206 g/mol. Express your a

castortr0y [2743]

Answer:

A) The molecular formula for ibuprofen is $C_{13}H_{18}O_2$

B) The molecular formula for Cadaverine is $C_{5}H_{14}N_2$

C) The molecular formula for Epinephrine is $C_9H_{13}O_3N_1$

Explanation:

Element percentage in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100$

A) The composition of ibuprofen, used for headaches, consists of 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by weight.

Ibuprofen has a molar mass of 206 g/mol.

The proposed molecular formula for ibuprofen is = $C_xH_yO_z$

Count of carbon atoms in one ibuprofen molecule;

$75.69\%=\frac{x\times 12 g/mol}{206 g/mol}\times 100$

$x=\frac{75.69\times 206 g/mol}{100\times 12 g/mol}=12.99\approx 13$

Count of hydrogen atoms in one ibuprofen molecule;

$8.80\%=\frac{y\times 1 g/mol}{206 g/mol}\times 100$

$y=\frac{8.80\times 206 g/mol}{100\times 1 g/mol}=18.12\approx 18$

Count of oxygen atoms in one ibuprofen molecule;

$15.51\%=\frac{z\times 16 g/mol}{206 g/mol}\times 100$

$z=\frac{15.51\times 206 g/mol}{100\times 16 g/mol}=1.99\approx 2$

Molecular formula for ibuprofen:

= $C_xH_yO_z= C_{13}H_{18}O_2$

B) Cadaverine consists of 58.55% carbon, 13.81% hydrogen, and 27.40% nitrogen by weight

Cadaverine has a molar mass of 102.2 g/mol.

The proposed molecular formula for Cadaverine is = $C_xH_yN_z$

Count of carbon atoms in one Cadaverine molecule;

$58.55\%=\frac{x\times 12 g/mol}{102.2 g/mol}\times 100$

$x=\frac{58.55\times 102.2 g/mol}{100\times 12 g/mol}=4.98\approx 5$

Count of hydrogen atoms in one Cadaverine molecule;

$13.81\%=\frac{y\times 1 g/mol}{102.2 g/mol}\times 100$

$y=\frac{13.81\times 102.2 g/mol}{100\times 1 g/mol}=14.11\approx 14$

Count of nitrogen atoms in one Cadaverine molecule;

$27.40\%=\frac{z\times 14 g/mol}{102.2 g/mol}\times 100$

$z=\frac{27.40\times 102.2 g/mol}{100\times 14 g/mol}=2.00\approx 2$

Molecular formula for Cadaverine:

= $C_xH_yN_z= C_{5}H_{14}N_2$

C) Epinephrine includes 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and 7.7% nitrogen by weight

Epinephrine has a molar mass of 180 g/mol.

The proposed molecular formula for Epinephrine is = $C_xH_yO_zN_w$

Count of carbon atoms in one Epinephrine molecule;

$59.0\%=\frac{x\times 12 g/mol}{180 g/mol}\times 100$

$x=\frac{59.0\times 180 g/mol}{100\times 12 g/mol}=8.85\approx 9$

Count of hydrogen atoms in one Epinephrine molecule;

$7.1\%=\frac{y\times 1 g/mol}{180 g/mol}\times 100$

$y=\frac{7.1\times 180 g/mol}{100\times 1 g/mol}=12.78\approx 13$

Count of oxygen atoms in one Epinephrine molecule;

$26.2\%=\frac{z\times 16 g/mol}{180 g/mol}\times 100$

$z=\frac{26.2\times 180 g/mol}{100\times 16 g/mol}=2.94\approx 3$

Count of nitrogen atoms in one Epinephrine molecule;

$7.7\%=\frac{w\times 14 g/mol}{180 g/mol}\times 100$

$w=\frac{7.7\times 180 g/mol}{100\times 14 g/mol}=0.99\approx 1$

Molecular formula for Epinephrine:

= $C_xH_yO_zN_w= C_9H_{13}O_3N_1$

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