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Taya2010 [7]
3 years ago
12

1. The three (3) main types of orbitals (s,p and d) depends on the__________quantum number.

Chemistry
1 answer:
vfiekz [6]3 years ago
3 0

Answer:

azimutal quantum number,

principal quantum number

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A certain first-order reaction 45% complete in 65seconds, determine the rate constant and the half life for the process ​
masha68 [24]

The rate constant : k = 9.2 x 10⁻³ s⁻¹

The half life : t1/2 = 75.3 s

<h3>Further explanation</h3>

Given

Reaction 45% complete in 65 s

Required

The rate constant and the half life

Solution

For first order ln[A]=−kt+ln[A]o

45% complete, 55% remains

A = 0.55

Ao = 1

Input the value :

ln A = -kt + ln Ao

ln 0.55 = -k.65 + ln 1

-0.598=-k.65

k = 9.2 x 10⁻³ s⁻¹

The half life :

t1/2 = (ln 2) / k

t1/2 = 0.693 : 9.2 x 10⁻³

t1/2 = 75.3 s

3 0
3 years ago
Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0
3 years ago
Which of the following is not an effect of acid rain? a. Corrosion of buildings c. Limestone corrosion b. Reduced cold tolerance
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<span> d. Reduced visibility

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3 years ago
Identify the correct equilibrium constant expression
SCORPION-xisa [38]

Answer: Keq= [CO2]^6[H2O]^6/ [O2]^6

Explanation:

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Volcanoes can destroy each and every thing
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