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Digiron [165]
3 years ago
9

The following equation shows the equilibrium in an aqueous solution of ammonia:NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Which of the follo

wing represents a conjugate acid-base pair?NH3 and H2ONH4+ and OH−H2O and OH−NH3 and
Chemistry
1 answer:
victus00 [196]3 years ago
4 0

Answer:

OH⁻ and H₂O

Explanation:

From the equation given in question ,

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq)

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

When an acid donates a proton, it changes into a base which is known as its conjugate base.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.

Hence , from the equation given in the question ,

The conjugate acid - base pair is -

OH⁻ and H₂O

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For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
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Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

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3 years ago
Suppose you have 100 grams of radioactive plutonium-239 with a half-life of 24,000 years. how many grams of plutonium-239 will r
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To solve this problem, let us first calculate for the rate constant k using the half life formula:

t1/2 = ln 2 / k

where t1/2 = half life period = 24,000 years, therefore k is:

k = ln 2 / 24,000

k = 2.89 x 10^-5 / yr

 

Now we use the rate equation:

A = Ao e^(-k t)

where,

A = mass of Plutonium-239 after number of years

Ao = initial mass of Plutonium-239

t = number of years

 

A. t = 12,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 12,000)

A = 70.7 g

 

B. t = 24,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 24,000)

A = 50 g

 

C. t = 96,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 96,000)

<span>A = 6.24 g</span>

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