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Degger [83]
2 years ago
5

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [

NO 2 ] drops from 0.0100 to 0.00650 M in 100 s. The rate of disappearance of NO2 for this period is __________ M/s.
Chemistry
1 answer:
serious [3.7K]2 years ago
6 0

Answer: The rate of disappearance of NO_2 is 3.5\times 10^{-5}M/s

Explanation:

The given chemical reaction is:

2NO_2\rightarrow 2NO+O_2

The rate of the reaction for disappearance of NO_2 is given as:

\text{Rate of disappearance of }NO_2=-\frac{\Delta [NO_2]}{\Delta t}

Or,

\text{Rate of disappearance of }NO_2=-\frac{C_2-C_1}{t_2-t_1}

where,

C_2 = final concentration of NO_2 = 0.00650 M

C_1 = initial concentration of NO_2 = 0.0100 M

t_2 = final time = 100 minutes

t_1 = initial time = 0 minutes

Putting values in above equation, we get:

\text{Rate of disappearance of }NO_2=-\frac{0.00650-0.0100}{100-0}\\\\\text{Rate of disappearance of }NO_2=3.5\times 10^{-5}M/s

Hence, the rate of disappearance of NO_2 is 3.5\times 10^{-5}M/s

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Answer: 0.0043mole

Explanation:Please see attachment for explanation

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2 years ago
Why does increasing the number of trials increase confidence in the results of the experiment?
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Match the specific enzyme to its class. A. oxidoreductase B. hydrolase C. transferase D. ligase E. lyase aminase
lilavasa [31]

Answer: Every enzyme has a specific name that can give us insight into the specific reaction that that enzyme can catalyze. We divide them into six different categories.

1) Oxidoreductase - includes two different types of reactions by transferring electrons from either molecule A to B or vice versa. It is involved in oxidizing electrons away from a molecule.

2) Hydrolase - uses water to divide a molecule into two other molecules.

3) Transferase - you move some functional group X from molecule B to molecule A

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5) Lyase - divides a molecule into two other molecules without using water and without reducing or oxidation

3 0
2 years ago
. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture
postnew [5]

Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

3 0
2 years ago
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alex41 [277]

Answer:

A

Explanation:

Let's illustrate this; see the attachment.

We see that Mrs. Jacobson is pushing to the right with a force of 100 N and there is another opposite force pushing with a force of 15 N. Since these are in opposite directions, we can say that the force opposite to Mrs. Jacobson is pushing the fridge -15 N to the right (instead of 15 N to the left).

The net force would then be:

100 N + (-15 N) = 85 N to the right

The answer is A.

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