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Shalnov [3]
3 years ago
5

Electron dot structure for Acetophenone??

Chemistry
1 answer:
Natali [406]3 years ago
3 0
Search it on the internet it’s easier
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Two students prepare lab reports one for a chemistry class and the other for a biology class which of the following items will b
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Answer:

the results of an investigation thank me later

Explanation:

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3 years ago
A sequence of thermonuclear fusion processes inside massive stars can continue to transform the nuclei of elements such as carbo
nlexa [21]

Answer:

Iron is the element that is produced at the limit of the reaction.

Explanation:

In nuclear fusion 2 lighter nuclei are combined together into a single nucleus releasing a tremendous amount of energy in the process.

Up to fusion of iron the reaction of fusion is exothermic but when iron atom pops out the reaction becomes endothermic and it requires very high amount of energy to fuse iron atoms which is not available thus marking an end to the fusion reaction.

5 0
2 years ago
Which of the followings is true about G0'? A. G0' can be determined using Keq' B. G0' indicates if a reaction can occur under no
nekit [7.7K]

Answer:

A and D are true , while B and F statements are false.

Explanation:

A) True.  Since the standard gibbs free energy is

ΔG = ΔG⁰ + RT*ln Q

where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R

when the system reaches equilibrium ΔG=0 and Q=Keq

0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)

therefore the first equation also can be expressed as

ΔG = RT*ln (Q/Keq)

thus the standard gibbs free energy can be determined using Keq

B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions

C) False. From the equation presented

ΔG⁰ = (-RT*ln Keq)

ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1

for example, for a reversible reaction  ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)

D) True. Standard conditions refer to

T= 298 K

pH= 7

P= 1 atm

C= 1 M for all reactants

Water = 55.6 M

5 0
3 years ago
Distinguish between atom molecules, and ions
NNADVOKAT [17]

Explanation:

Atoms are the smallest unit of matter that can't be broken down chemically. Molecules are groups of two or more atoms that are chemically bonded. Ions are atoms or molecules that have gained or lost one or more of their valence electrons and therefore have a net positive or negative charge.

7 0
3 years ago
Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc
Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is C_5H_7N

(b) The empirical formula for the given compound is C_4H_5N_2O

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = \frac{6.17}{1.23}=5.01\approx 5

For Hydrogen  = \frac{8.70}{1.23}=7.07\approx 7

For Nitrogen = \frac{1.23}{1.23}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is C_5H_7N_1=C_5H_7N

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = \frac{4.12}{1.03}=4

For Hydrogen  = \frac{5.19}{1.03}=5.03\approx 5

For Nitrogen = \frac{2.06}{1.03}=2

For Nitrogen = \frac{1.03}{1.03}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is C_4H_5N_2O_1=C_4H_5N_2O

6 0
3 years ago
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