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3 years ago

How many moles of O2 are produced in the decomposition of 24 g of water in the

1 answer:

8 0

M of water = 2(1) + 16 = 18 g/mol
n of water = 24g/(18g/mol) = 4/3 mol
n of H2O : n of O2 = 2 : 1
2:1 = 4/3 : x
x = (4/3)/2 = 2/3 mol of O2

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The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i

Rus_ich [418]

Answer: a) The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) concentration after 8.8 min:

$8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}$

$\log\frac{0.25}{a-x}=0.15$

$\frac{0.25}{a-x}=1.41$

$(a-x)=0.17M$

b) for concentration to decrease from 0.25M to 0.15M

$t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20$

$t=687s$

7 0

3 years ago

Assuming all volume measurements are made at the same temperature and pressure, how many liters of oxygen gas would it take to r

EleoNora [17]

2H2 (g) + O2 (g) -->2H2 O(g)
mole ratio of H2:O2=2:1
7.25/2=3.625

3 0

3 years ago

Which of the following is an example of improved functioning as related to the benefits of muscular strength and endurance train

Evgen [1.6K]

I believe the answer is letter A

7 0

3 years ago

Read 2 more answers

Several general characteristics of metabolic reactions or reaction pathways are listed below. sort them into the appropriate met

Kobotan [32]

Answers are:
Catabolism:
- generally exergonic (spontaneous): In this reactions energy is released.
- convert NAD+ to NADH. Electrons and protons released in reactions are attached to NAD+.
- generation of ATP. ATP is synthesis from ADP.
- convert large compounds to smaller compounds. Foe example starch to monosaccaharides.
Anabolism:
- convert NADPH to NADP+. Protons and electrons are used to make chemical bonds.
- convert small compounds to larger compounds.

8 0

3 years ago

Read 2 more answers

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$

5 0

3 years ago