Do chemical equations with no reaction still have to be balanced

Chemistry

1 answer:

dangina [55]3 years ago

5 0

Answer:

The same number of atoms of each element must appear on both sides of a chemical equation. However, simply writing down the chemical formulas of reactants and products does not always result in equal numbers of atoms. You have to balance the equation to make the number of atoms equal on each side of an equation.

Explanation:

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Your friend challenged you to a race you know in order to beat him you must run 15 meters within 20 seconds in the northern dire

Marat540 [252]

Answer:

.75 meter north

Explanation:

v = d/t

v = 15/20

v = .75

7 0

3 years ago

This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): An analytical chemist has det

jeyben [28]

Answer:

0.054 mol O

Explanation:

<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of carbon in a sample of acetic acid. How many moles of oxygen are in the sample?</em>

<em />

Step 1: Given data

Chemical formula of acetic acid: CH₃CO₂H

Moles of carbon in the sample: 0.054 moles

Step 2: Establish the appropriate molar ratio

According to the chemical formula, the molar ratio of C to O is 2:2.

Step 3: Calculate the moles of oxygen in the sample

We will use the molar ratio to determine the moles of oxygen accompanying 0.054 moles of carbon.

0.054 mol C × (2 mol O/2 mol C) = 0.054 mol O

6 0

3 years ago

Products: 1-methylcyclohexene, 3-methylcyclohexene, methylenecyclohexane

Lostsunrise [7]

Answer:

See explanation

Explanation:

The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.

Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.