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balandron [24]
3 years ago
14

The reaction NO2(g) + CO(g) → NO(g) + CO2(g) has been found to be second order with respect to NO2 and zero order with respect t

o CO. At a certain temperature, the rate constant is found experimentally to be 3.0 × 10−5 L mol · s . What is the rate of formation
Chemistry
1 answer:
hodyreva [135]3 years ago
7 0

Answer:

The answer is " 122.88 \times 10^{-5}"

Explanation:

In the given question some data is missing. so, the correct solution can be defined as follows:

Missing values:

k= 3.0 \times 10^{-5}\\\\NO_2 =6.4 \ mol / l \\\\CO= 4.1 mol / l \\\\NO= 5.2 mol / l \\\\CO_2= 1.3 mol / l\\

Given equation:

NO_2(g) + CO(g) \longrightarrow  NO(g) + CO_2(g)

In the above equation, the rate is =k.[NO_2]^2 because the above given is the part of the second-order, which relates to NO_2. In the zeros order the Carbon monoxide (CO) its reaction doesn't affect the rate.

Calculating the rate:

Rate =k.[NO_2]^2

        = 3.0 \times 10^{-5} \times 6.4^2\\\\=3.0 \times 10^{-5} \times 40.96\\\\= 122.88 \times 10^{-5}\\

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Which best explains why ionization energy tends to decrease from the top to the bottom of a group?
Setler79 [48]

Answer:

Electrons are far apart from the nucleus as we move down the group.

Explanation:

The ionization energy is the amount of energy which is necessary to remove an electron from an atom.

In an atom there exist a force of attraction at the center (nucleus). This is because of the positive charge which exists in the nucleus. This force of attraction is less felt as the distance between the electron and the proton increases. Hence the ionization energy increases as the number of shells increases for an atom. As we move down the group in the periodic table, the number of shells increases which implies a decrease in ionization energy.

5 0
3 years ago
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Find the [H+] in an acetic acid solution that has a pH of 5.12
Eva8 [605]
<span>Data:
pH = 5.2
[H+] = ?

Knowing that: (</span><span>Equation to find the pH of a solution)</span>
pH = -log[H+]
<span>
Solving:
</span>pH = -log[H+]
5.2 = - log [H+]
Knowing that the exponential is the opposite operation of the logarithm, then we have:
[H+] = 10^{-5.2}
\boxed{\boxed{[H+] = 6.30*10^{-6}}}\end{array}}\qquad\quad\checkmark
3 0
2 years ago
What is an example of entropy from everyday life?
Lorico [155]
One of the greatest examples of entropy working in everyday life is the fact that a messy room is the norm. So if you clean your room it will likely become messy over a period of time, and work will have to be put in to make it orderly again.
4 0
2 years ago
Chlorine has two isotopes, 35Cl and 37Cl; 75.77 % of chlorine is 35Cl and 24.23 % is 37Cl. The atomic mass of 35Cl is 34.969 amu
storchak [24]

Answer:

35.4528731 amu

Explanation:

To appropriately get the atomic mass unit of chlorine, we can get the answer using the masses from the isotopes. This can be obtained as follows. What we do is that we multiply the percentage compositions by the masses.

Now let’s do this.

[75.77/100 * 34.969] + [24.23/100 * 36.966]

= 26.4960113 + 8.9568618 = 35.4528731

3 0
3 years ago
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A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
2 years ago
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