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3 years ago

Balance this equation. if a coefficient of "1" is required, choose "blank" for that box.

1 answer:

7 0

Step 1: Write Imbalance Equation

CH₃CHO + O₂   → CO₂ + H₂O

Step 2: Balance Carbon Atoms:
There are 2 carbon atoms at reactant side and one at product side. So multiply CO₂ with 2 to balance them. i.e.

CH₃CHO + O₂   → 2 CO₂ + H₂O

Step 3: Balance Hydrogen Atoms:
There are 4 hydrogen atoms at reactant side and 2 Hydrogen atoms at product side. So, multiply H₂O by 2 to balance Hydrogen on both sides. i.e.

  CH₃CHO + O₂   → 2 CO₂ + 2 H₂O

Step 4: Balance Oxygen Atoms:
There are 3 Oxygen atoms at reactant side and 6 Oxygen atoms at product side. In order to balance them multiply O₂ on reactant side by 2.5 (5/2). i.e

CH₃CHO + 5/2 O₂   →   2 CO₂ +  2 H₂O

Step 6: Eliminate Fraction:
Multiply overall equation by 2 to eliminate fraction. i.e.

2 CH₃CHO + 5 O₂   →   4 CO₂ +  4 H₂O

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Which of the following substances would you predict to have the highest DHvap? Group of answer choices CH3CH2CH2CH3 CH3CH2OH HF

Katyanochek1 [597]

Answer:

a) $CH_3CH_2CH_2CH_3$

Explanation:

In this question we have the following answer choices:

a) $CH_3CH_2CH_2CH_3$

b) $CH_3CH_2OH$

c) $HF$

d) $CH_3Cl$

e) $HOCH_2CH_2OH$

We have to remember the relationship between intermolecular forces and vapor pressure. If we have stronger intermolecular forces we will have less vapor pressure because the molecules have more interactions between them, so, the molecules will prefer to stay in a liquid state rather than a gaseous state. Now, we have to check each molecule:

a) $CH_3CH_2CH_2CH_3$ (Van der waals interactions)

b) $CH_3CH_2OH$ (Hydrogen bonding)

c) $HF$ (Hydrogen bonding)

d) $CH_3Cl$ (Dipole-dipole interaction)

e) $HOCH_2CH_2OH$ (Hydrogen bonding)

For molecules b, c and e we have hydrogen bond to a heteroatom (O, N, S, or P). In this case oxygen, therefore we will have hydrogen bonding interactions (a very strong interaction). So, we can discard these ones.

In molecule e, we have "Cl" bond to a "C" therefore we will have the presence of a dipole (due to the electronegativity difference). If we have a dipole, we will have a dipole-dipole interaction (a strong interaction, less than hydrogen bonding but still is a strong interaction).

In molecule a, we have only Van der Waals interactions because in this molecule we have only carbon and hydrogen atoms bonded by single bonds. So, we will have a non-polar molecule. These interactions are the weakest interactions of all the molecules given. So, if we have weaker interactions the molecules can be converted to a gas state more easily and we have more vapor pressure. 

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3 years ago

Why do we use acid during the detection of halogens

ohaa [14]

Answer:

Nitric acid decomposes sodium cyanide and sodium halide. else, they precipitate in test and misguide the result. Therefore, dilute nitric acid is added before testing halogens to expel all the gases if evolved.

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2 years ago

Arrange the bonds between H, O, and F from most polar to least polar based on electronegativity.

Andrews [41]

Answer:

H-F> H-O> O-F

while H-H is nonpolar.

Explanation:

The bond is polar when electronegativity difference of both bonded atom is greater than 0.4.

Electronegativity values of O. F and H are

O = 3.4

H = 2.2

F = 3.98

For O-F

3.98 - 3.4 = 0.58

For H-F

3.98 - 2.2 = 1.78

For H-O

3.4 - 2.2 = 1.2

For H-H

2.2 -2.2 = 0

So H-H is nonpolar.

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3 years ago

What does an electromagnetic spectrum refer to?

Juliette [100K]

The entire distribution of electromagnetic radiation according to frequency or wavelength

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3 years ago

The atomic symbol superscript 40 subscript 19 upper k. represents potassium-40 (k-40), a radioactive isotope that has 19 protons

Oksana_A [137]

Answer:

40/18AR

Explanation:

sorry my answer was late but i hope this helps someone else! I got it right on the test

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2 years ago