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Deffense [45]
2 years ago
10

Choose the options below that are true.

Chemistry
1 answer:
Assoli18 [71]2 years ago
5 0

Answer:

The options <u>(A) -</u>The rate law for a given reaction can be determined from a knowledge of the rate-determining step in that reaction's mechanism.  and <u>(C) </u>-The rate laws of bimolecular elementary reactions are second order overall ,<u>is true.</u>

Explanation:

(A) -The rate law can only be calculated from the reaction's slowest or rate-determining phase, according to the first sentence.

(B) -The second statement is not entirely right, since we cannot evaluate an accurate rate law by simply looking at the net equation. It must be decided by experimentation.

(C) -Since there are two reactants, the third statement is correct: most bimolecular reactions are second order overall.

(D)-The fourth argument is incorrect. We must track the rates of and elementary phase that is following the reaction in order to determine the rate.

<u>Therefore , the first and third statement is true.</u>

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Look at the potential energy diagram for a chemical reaction.
vladimir1956 [14]

Answer: The activation energy is 10 kJ and the reaction is exothermic.

Explanation: Exothermic reactions are those in which heat is released and thus the energy of products is less than the energy of reactants.

Endothermic reactions are those in which heat is absorbed and thus the energy of products is more than the energy of reactants.

Activation energy is the extra amount of energy required by the reactants to cross the energy barrier to convert to products.

Given : Energy of reactants = 40kJ

Energy of activation: (50-40)=10 kJ

Energy of products = 50 kJ

Energy of products = 15 kJ

Thus Energy of products (15kJ) < Energy of reactants(40kJ), the reaction is exothermic as energy has been lost to surroundings in the form of heat.

8 0
3 years ago
An atomic number stands for the number of _____. neutrons in the nucleus of an atom protons in the nucleus of an atom valence el
Effectus [21]

Answer:

atoms or electrons

Explanation:

but l guess electrons is the best answer

5 0
3 years ago
What happens to electrons in the photoelectric effect?
11111nata11111 [884]
The photoelectric effect occurs when light shines on a metal. ... Light of any frequency will cause electrons to be emitted.
5 0
2 years ago
Read 2 more answers
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

Now put all the given values in this expression, we get:

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

Now put all the given values in this expression, we get:

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
3 years ago
A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula
Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

= 2.857 x 10^14 g of gold in the ocean.

1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

= $5,840,000,000,000,000

8 0
3 years ago
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