Answer:
0.6743 M
Explanation:
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:
- Molarity = moles / volume
- moles = Molarity * volume
- 0.4293 M * 39.27 mL = 16.86 mmol NaOH
<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.
Finally we <u>calculate the concentration (molarity) of acetic acid</u>:
- 16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M
<u>Answer:</u> 2.00 atm
<u>Explanation:</u>
The gas is kept under the same temperature in this problem. Assuming the amount of gas is constant, we can apply the Boyle's law.
The Boyle's law equation,
P₁V₁ = P₂V ₂
Plug in the values,
1.00 atm x 4.0 L = P₂ x 2.0 L
Simplify,
4.00 atm L = 2 P₂ L
Now flip the equation,
2 P₂ L = 4.00 atm L
Dividing both sides by 2 we get,
P₂ = 2.00 atm
Answer:
b Different amounts of food samples were used.
Explanation:
The mass of the two samples needs to be the same in order for the test to be accurate.
Complete question:
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
See attached image for the structure formula of the compound
Answer:
(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane
Explanation:
If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.
Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).
The condensed formula will be written as;
(CH₃)₂CHC(CH₃)₃
This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane
The H+ concentration that would best describe a basic solution would be the one having values less than 10^-7. The pH of a solution is related to H+ concentration by pH = -log[H+]. Therefore, as the concentration of H+ decreases the alkalinity would rise.
