It can be done. Normally the boiling point of water is 100°C. It will boil at temperature greater than 100°C more quickly. Water can be boiled at 95°C but for that the atmospheric pressure of the water should be decreased which will decrease the boiling point of water.
<h3>
Concept :</h3>
To boil water at 95°C, decrease the atmospheric pressure.
At 105°C, the water will be boiling quickly than normal at 100°C.
First, we calculate of the concentration of the H+ ions in the solution from the pH given. Then, calculate the new concentration after dilution. Calculation are as follows:
pH = -log[H+]
5 = -log[H+]
[H+] = 1 x 10^-5 M
M1V1 = M2V2
<span>1 x 10^-5 M (V1) = M2(100V1)
</span>M2 = 1 x 10^-7
pH = -log[<span>1 x 10^-7</span>]
pH = 7
The equation : y=3x-5
<h3>Further explanation
</h3>
Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates
General formula
y-y1 = m(x-x1)
or
y = mx + c
Where
m = straight-line gradient which is the slope of the line
x1, y1 = the Cartesian coordinate that is crossed by the line
c = constant
The formula for a gradient (m) between 2 points in a line
m = Δy / Δx


Answer:
= 9.872002 × 10^6
Explanation:
Move the decimal point in your number until there is only one non-zero digit to the left of the decimal point. The resulting decimal number is a.
Count how many places you moved the decimal point. This number is b.
If you moved the decimal to the left b is positive.
If you moved the decimal to the right b is negative.
If you did not need to move the decimal b = 0.
Write your scientific notation number as a x 10^b and read it as "a times 10 to the power of b."
Remove trailing 0's only if they were originally to the left of the decimal point.
The balanced reaction for combustion is as follows ;
2C₂H₅OH + 6O₂ ---> 4CO₂ + 6H₂O
the stoichiometry of C₂H₅OH to O₂ is 2:6
that means 2 mol of C₂H₅OH reacts with 6 mol of O₂.
when 1 mol of C₂H₅OH reacts with 6/2 mol of O₂,
then 0.3020 mol of C₂H₅OH reacts with - 6/2 x 0.3020
therefore number of O₂ moles reacted = 0.91 mol