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ddd [48]
2 years ago
15

After a certain pesticide compound is applied to crops, its decomposition is a first-order reaction with a half-life of 56 days.

What is the rate constant, k, for the decomposition reaction?
Chemistry
1 answer:
Anna35 [415]2 years ago
8 0

The rate constant, k, for the decomposition reaction :  k = 0.0124 / days

<h3>Further explanation</h3>

Given

The half-life of 56 days

Required

The rate constant, k

Solution

For first-order, rate law : ln[A]=−kt+ln[A]o

The half-life :  the time required to reduce to half of its initial value.

The half life :

t1/2 = (ln 2) / k

k = (ln 2) / t1/2

k = 0.693 / 56 days

k = 0.0124 / days

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One ATP molecule's hydrolysis releases 7.3 kcal/mol of energy (∆G = −7.3 kcal/mol of energy). If it takes 2.1 kcal/mol of energy
Oxana [17]

Answer:

One ATP molecule's hydrolysis can move 3 ions of Sodium (Na+) across the menbrane.

Explanation:

As you can see, the energy provided by ATP is enough for moving 3 ions of Na+. Each ion needs +2.1 kcal/mol of energy.

If we multiply by 3 the energy for moving across the membrane= +6.3 kcal/mol

By adding the energy from ATP:

ΔGTotal=6.3-7.3= -1 kcal/mol

7 0
2 years ago
Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T
Llana [10]

Answer:

A

Explanation:

Iron has the ground state electronic configuration [Ar]3d64s2

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3 0
2 years ago
 A reaction container holds 5.77 g of P4 and 5.77 g of O2.
Dvinal [7]

Answer:

a) O2 is the limiting reactant

b) 5.75 grams P4O10

c) 5.79 grams P4O6

Explanation:

Step 1: Data given

Mass of P4 = 5.77 grams

Mass of O2 = 5.77 grams

Molar mass of P4 = 123.90 g/mol

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

P4 + 3O2 → P4O6

Step 3: Calculate moles of P4

Moles P4 = mass P4 / molar mass P4

Moles P4 = 5.77 grams / 123.90 g/mol

Moles P4 = 0.0466 moles

Step 4: Calculate moles O2

Moles O2 = mass O2 / molar mass O2

Moles O2 = 5.77 grams / 32.0 g/mol

Moles O2 = 0.1803 moles

Step 5: Calculate limiting reactant

P4 is the limiting reactant in this reaction. It will completely be consumed (0.0466 moles). O2 is in excess, there will react 3*0.0466 = 0.1398 moles

There will remain 0.1803 - 0.1398 = 0.0405 moles O2

Step 6: Calculate the amount of P4O6

For 1 mol P4 we'll have 1 mol P4O6

For 0.0466 moles P4 we'll have 0.0466 moles P4O6

Step 7: The balanced equatio

P4O6 + 2O2 → P4O10

We have 0.0466 moles P4O6 and 0.0405 moles O2

Step 8: Calculate the limiting reactant

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

O2 is the limiting reactant. It will completely be consumed (0.0405 moles)

P4O6 is in excess. There will react 0.0405/2 = 0.02025 moles

There will remain 0.0466 - 0.02025 = 0.02635 moles P4O6

This is 0.02635 * 219.88 g/mol = 5.79 grams P4O6

Step 9: Calculate moles and mass of P4O10

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

For 0.0405 moles O2 we'll have 0.02025 moles P4O10

This is 0.02025 * 283.89 g/mol = 5.75 grams P4O10

3 0
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ExtremeBDS [4]
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pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid]) 
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3 0
2 years ago
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