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almond37 [142]
1 year ago
7

The rate law of the reaction NH3 + HOCl → NH2Cl + H2O is rate = k[NH3][HOCl] with k = 5.1 × 106 L/mol·s at 25°C. The reaction is

made pseudo-first-order in NH3 by using a large excess of HOCl. How long will it take for 40% of the NH3 to react if the initial concentration of HOCl is 2 × 10−3 M?
Chemistry
1 answer:
SIZIF [17.4K]1 year ago
6 0

Answer:

40% of the ammonia will take 4.97x10^-5 s to react.

Explanation:

The rate is equal to:

R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]

R = k´ * [NH3]

k´ = 10200 s^-1

Because k´ is the psuedo first-order rate constant, we have the following:

b/(b-x) = 100/(100-40) ; 40% ammonia reacts

b/(b-x) = 1.67

log(b/(b-x)) = log(1.67)

log(b/(b-x)) = 0.22

the time will equal to:

t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s

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deff fn [24]

Answer:

The correct answer is "32%".

Explanation:

The given values:

Weight of H,

= 4.9 g

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= 15.8 g

Now,

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= \frac{Weight \ of \ C}{Total \ weight}\times 100

By substituting the values, we get

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3 0
2 years ago
Approximately what volume do 3.0 moles of kr gas occupy at stp? 80.0 40.0 l none of the above 20.0
podryga [215]
None of the above.
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4 0
2 years ago
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Write a balanced half-reaction for the oxidation of liquid water H2O to aqueous hydrogen peroxide H2O2 in basic aqueous solution
nignag [31]

Answer : The balanced half-reaction in a basic solution will be,

2OH^-(aq)\rightarrow H_2O_2(aq)+2e^-  

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion (OH^-) at that side where the less number of hydrogen are present.

Now balance the charge.

  • The half reaction is :

H_2O(l)\rightarrow H_2O_2(aq)

  • Now balance the oxygen atoms.

H_2O(l)\rightarrow H_2O_2(aq)+H_2O(l)

  • Now balance the hydrogen atoms.

H_2O(l)+2OH^-(aq)\rightarrow H_2O_2(aq)+H_2O(l)

  • Now balance the charge.

H_2O(l)+2OH^-(aq)\rightarrow H_2O_2(aq)+H_2O(l)+2e^-

The balanced half-reaction in a basic solution will be,

2OH^-(aq)\rightarrow H_2O_2(aq)+2e^-

7 0
2 years ago
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r
Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

6 0
2 years ago
Starting with 1.5052g of BaCl2•2H2O and excessH2SO4, how many grams of BaSO4 can be formed?
muminat
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba 
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
6 0
2 years ago
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