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garri49 [273]
3 years ago
7

Can you answer both questions please??

Chemistry
1 answer:
zhenek [66]3 years ago
3 0
The air we breathe is a mixture and for number 10th they are elements.
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A reaction is exothermic when Group of answer choices weak bonds break and strong bonds form. strong bonds break and weak bonds
Leona [35]

Answer:

weak bonds break and strong bonds form

Explanation:

An exothermic reaction is a chemical reaction in which heat energy is evolved during the reaction process.

Bond formation requires energy while bond breakage releases energy. More energy is needed for the formation of weak bonds as compared to strong bonds.

<em>Hence, when weak bonds break, they release more energy than needed to make a corresponding strong bond leading to the release of the remaining energy as heat.</em>

6 0
3 years ago
Describe how the surface area affect the rate of a reaction
leva [86]
When we increase the surface area of an object, more atoms are exposed. Since more atoms are exposed, the atoms can react faster, and this is why the rate of a reaction increases when the surface area increases.

For example, lets say we want to heat a potato. If we just put the whole potato in the microwave, it will take a long time for the potato to get thoroughly heated. However, if we chop the potato into smaller pieces, we will observe that it gets heated much faster. This is because we increased the surface area of the potato, which resulted in more potato atoms to be exposed to the heat, and caused the reaction to be faster.
7 0
3 years ago
The incomplete table below shows selected properties of compounds that have ionic, covalent, or metallic bonds.
pogonyaev

Answer:

is there an option for silver? if so, silver is the answer.

8 0
1 year ago
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
In the haber process, ammonia is synthesized from nitrogen and hydrogen: n2 (g) + 3h2 (g) → 2nh3 (g) δg° at 298 k for this react
allsm [11]

Answer:

e- 7.25 x 10³.

Explanation:

∵ ΔG = -RTlnK,

where, ΔG is the free energy change.

R is the general gas constant (R = 8.324 J/mol.K).

K is the equilibrium constant of the reaction.

  • For the reaction: <em>N₂(g) + 3H₂(g) → 2NH₃(g),</em>

K = (PNH₃)²/(PN₂)(PH₂)³ = (0.65)²/(1.9)(1.6)³ = 5.43 x 10⁻².

∵ ΔG = -RTlnK.

∴ ΔG = -(8.314 J/mol.K)(298 K) ln(5.43 x 10⁻²) = 7.218 x 10³ J/mol.

3 0
3 years ago
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