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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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A solution contains 0.0150 M Pb2+(aq) and 0.0150 M Sr2+(aq) . If you add SO2−4(aq) , what will be the concentration of Pb2+(aq)
AnnyKZ [126]

Answer:

\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}

Explanation:

1. Concentration of SO₄²⁻

SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷

                   0.0150          x

K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}

2. Concentration of Pb²⁺

PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸

                        x          2.293 × 10⁻⁵

K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}

 

4 0
1 year ago
The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]

In NH3 , let oxidation number of N be x

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In HNO3 , let oxidation number of N be x

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2 years ago
identify A and B, isomers of molecular formula C3H4Cl2, from the given 1H NMR data: Compound A exhibits peaks at 1.75 (doublet,
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Answer:

See explaination

Explanation:

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2 years ago
Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe
ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of CH_4 will be,

2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=29.7kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=-41kJ

(3) CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206kJ

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3

\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)

\Delta H_{rxn}=-187.6kJ/mol

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

\Delta H=\frac{q}{n}

or,

q=\Delta H\times n

where,

\Delta H = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = \frac{1.00\times 1000g}{12.00g/mol}=83.33mol

Now put all the given values in the above formula, we get:

q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ

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