1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
arlik [135]
3 years ago
15

Water can be formed in the following reaction:

Chemistry
1 answer:
andreev551 [17]3 years ago
6 0

Answer is: 10 moles of water will be produced.

Balanced chemical reaction of formation of water:

2H₂ + O₂ → 2H₂O.

n(H₂) = 10 mol; amount of hydrogen gas.

From balanced chemical reaction: n(H₂) : n(H₂O) = 2 : 2 (1 : 1).

n(H₂O) = n(H₂).

n(H₂O) = 10 mol; amount of water.

You might be interested in
A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.
Reika [66]

Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ /  2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

4 0
3 years ago
Directions: Perform the Assessment in Module 7 because that will serve as your Performance Task for Module 5, 6 & 7. Kindly
garik1379 [7]

please put your question completely....

8 0
3 years ago
Plsss help <br> will mark BRAINLIEST
satela [25.4K]

Answer:

c

Explanation: just a chemachal

5 0
3 years ago
What is the Law of Definite Composition and how does it apply to pure substances and mixtures?
Lyrx [107]

Answer:

The Law of Definite Proportions ensures that chemical compounds are always created using the same proportions, regardless of the amount of the compound which is being made

5 0
3 years ago
Read 2 more answers
The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment
astraxan [27]

Answer:

The activation energy is =8.1\,kcal\,mol^{-1}

Explanation:

The gas phase reaction is as follows.

A \rightarrow B+C

The rate law of the reaction is as follows.

-r_{A}=kC_{A}

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = 10dm^{3}

Temperature "T" = 300 K

Volumetric flow rate of the reaction v_{o}=5dm^{3}s

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]

Rearrange the formula is as follows.

k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)

The feed has Pure A, mole fraction of A in feed y_{A_{o}} is 1.

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacte.

=1(2-1)=1

Substitute the all given values in equation (1)

k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}

Therefore, the rate constant in case of the plug flow reacor at 300K is1.2s^{-1}

The rate constant in case of the CSTR can be calculated by using the formula.

\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)

The feed has 50% A and 50%  inerts.

Hence, the mole fraction of A in feed y_{A_{o}} is 0.5

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacted.

=0.5(2-1)=0.5

Substitute the all values in formula (2)

\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}

Therefore, the rate constant in case of CSTR comes out to be 2.8s^{-1}

The activation energy of the reaction can be calculated by using formula

k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}

Substitute the all values.

=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}

=8.1\,kcal\,mol^{-1}

Therefore, the activation energy is =8.1\,kcal\,mol^{-1}

8 0
3 years ago
Other questions:
  • Which phrase describes an atom
    5·1 answer
  • Does the identity of gas matter when predicting its behavior why or why not?
    7·1 answer
  • Which substance acts as the anode in the lead-acid automobile battery?
    15·1 answer
  • LiNO3(aq) and Na2CO3(aq).If a solid forms when solutions containing these salts are mixed, write the ionic equation.
    5·2 answers
  • The coinage metals -- copper, silver, and gold -- crystallize in a cubic closest packed structure. Use the density of silver (10
    15·1 answer
  • The lighted half of the moon faces away from Earth during the _____ moon phase.
    15·1 answer
  • What is exchanged between the system and the surroundings ​
    7·1 answer
  • Two plastic containers are washed in a dishwasher. One keeps its shape and the other becomes deformed.
    12·1 answer
  • Help please I really really need it now ​
    9·1 answer
  • What kind of weather is associated with high air pressure?
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!