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Rama09 [41]
2 years ago
11

Which of the following statements is true concerning the reaction given below?2H2S(g) + O2(g) → 2S(s) + 2H2O(g)a. The reaction i

s second-order in H2S(g) and first-order in O2(g).b. The reaction is first-order in H2S(g) and second-order in O2(g).c. The rate law is Rate = k[H2S]2[O2].d. The rate law is Rate = k[H2S][O2].e. The rate law may be determined only by experiment.
Chemistry
1 answer:
pantera1 [17]2 years ago
8 0

Answer:

The rate law may be determined only by experiment.

Explanation:

For a reaction, A + B ---> C, the rate law can only be determined from experimental data. Chemists determine the rate of reaction by carefully observing the changes in the concentration of species as the reaction progresses.

Hence, the rate law is not determined by inspection of the chemical reaction equation, it must be obtained from the experimental data, hence the answer given.

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During a laboratory experiment, 36.12 grams of Al2O3 was formed when O2 reacted with aluminum metal at 280.0 K and 1.4 atm. What
photoshop1234 [79]

Answer:

8.70 liters

Explanation:

4 0
2 years ago
Activity 1: My Incompleteness, Complete Me. Fill in the Punnett squares with the correct genotypes based on the key pictures tha
Xelga [282]

hope you can read it. hope it's right too :).

All the best.

8 0
2 years ago
When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the result­ing concentration of Na+ is 2
tiny-mole [99]

Answer : The resulting concentration of Na^+ ion is, 4.5 M

Explanation : Given,

Concentration of Na_2CO_3 = M_1 = 3.0 M = 3.0 mol/L

Volume of Na_2CO_3 = V_1 = 70 mL = 0.07 L

Concentration of NaHCO_3 = M_2 = 1.0 M = 1.0 mol/L

Volume of NaHCO_3 = V_2 = 30 mL = 0.03 L

First we have to calculate the moles of Na_2CO_3 and NaHCO_3

\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol

and,

\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol

Now we have to calculate the moles of Na^+ ions.

As, 1 mole of Na_2CO_3 will give 2 moles of Na^+ ions

So, 0.21 moles of Na_2CO_3 will give 2\times 0.21=0.42 moles of Na^+ ions

and,

As, 1 mole of NaHCO_3 will give 1 mole of Na^+ ions

So, 0.03 moles of NaHCO_3 will give 0.03 moles of Na^+ ions

So,

Total number of moles of Na^+ ions = 0.42 + 0.03 =0.45 mole

Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L

Now we have to calculate the concentration of Na^+ ions.

\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M

Therefore, the resulting concentration of Na^+ ion is, 4.5 M

6 0
2 years ago
Each student in a class placed a 2.00 g sample of a mixture of Cu and Al in a beaker and placed the beaker in a fume hood. The s
Aleks [24]

Answer:

Percentage mass of copper in the sample = 32%

Explanation:

Equation of the reaction producing Cu(NO₃) is given below:

Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)

From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.

Molar mass of copper = 64 g/mol

mass of copper = number of moles * molar mass

mass of copper = 0.01 mol * 64 g/mol = 0.64 g

Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%

Percentage mass of copper in the sample = 32%

3 0
2 years ago
Consider the following hypothetical reaction: 2 P + Q → 2 R + S The following mechanism is proposed for this reaction: P + P Q
aleksandr82 [10.1K]

Answer: the answer is option (D). k[P]²[Q]

Explanation:

first of all, let us consider the reaction from the question;

2P + Q → 2R + S

and the reaction mechanism for the above reaction given thus,

P + P ⇄ T     (fast)

Q + T → R + U    (slow)

U → R + S    (fast)

we would be applying the Rate law  to determine the mechanism.

The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.

It is important to understand that laws based on experiment do not allow for intermediate concentration.  

The mechanism steps for the reactions in the question  are given below when we add them by cancelling the intermediates on the opposite side of the equations then we get the overall reaction equation.

adding this steps gives a final overall reaction reaction.

2P + Q ------------˃ 2R + S

Thus the rate equation is given as

Rate (R) = K[P]²[Q]

cheers, i hope this helps

3 0
2 years ago
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