Answer:
Scientist used models to explain and predict the behavior of real object or system
A water solution is found to have a molar oh- concentration of 3.2 x 10-5. the solution would be classified as neutral.
The concentration of hydroxide ions (OH-) is measured by pOH. It is a way of expressing how alkaline a solution is. At 25 degrees Celsius, aqueous solutions with pOH values of 7 or less are neutral, whereas those with pOH values of 7 or more are acidic. The hydrogen ion potential is known as pH. The potential of hydroxide ions is known as pOH. 2. It is a scale used to estimate the hydrogen ion (H+) concentration in the solution. The hydroxide ion (OH-) concentration of the solution is measured using this scale.
pH + pOH = 14
pOH = 3.2x 10-5
[OH-] = 10^(-pOH) =10^(- 3.2x 10-5)
= 0.99
Answer:
hello your question is incomplete below is the missing part of the question
answer : 104°c
Explanation:
The Eutectic temperature for the mixture is 104°c
From the chart attached below it can be seen that the temperature from the two lines of best fit cross is 104°c
Three groups
gases, metals, metalliods/nonmetals
According to the balanced equation of the reaction:
2C2H2 + 5O2 → 4CO2 + 2H2O
So we can mention all as liters,
A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O
So, when we have 75L of CO2
and when we have 2 L of C2H2 reacts and gives 4 L of CO2
2C2H2 → 4CO2
∴ The volume of C2H2 required is:
= 75L / 2
= 37.5 L
B) and, when we have 75 L of CO2
and 4CO2 → 2H2O
∴ the volume of H2O required is:
= 75 L /2
= 37.5 L
C) and from the balanced equation and by the same way:
when 5 liters O2 reacts to give 4 liters of CO2
and we have 75 L of CO2:
5 O2 → 4 CO2
?? ← 75 L
∴ the volume of O2 required is:
= 75 *(5/4)
= 93.75 L
D) about the using of the number of moles the answer is:
no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.