100 times less H+
A solution at ph 10 contains<u> </u><u>100 times less H+</u> than the same amount of solution at ph 8.
<h3>The pH scale: How does it function?</h3>
- The pH scale determines how acidic or basic water is.
- The range is 0 to 14, with 7 representing neutrality.
- Acidity is indicated by pH values below 7, whereas baseness is shown by pH values above 7.
- In reality, pH is a measurement of the proportion of free hydrogen and hydroxyl ions in water.
<h3>How does the pH change when two acids are combined?</h3>
- An acid's strength increases with the quantity of hydrogen ions it releases.
- The pH of the strong acids is between 1 and 2.
- We may observe that there is no response when two acids of the same strength are combined.
- It's because the end product will be neutral and the pH won't change.
<h3>How is pH value determined?</h3>
There are two ways to measure pH:
- colorimetrically with indicator fluids or sheets
- electrochemically with electrodes and a millivoltmeter for greater accuracy (pH meter).
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Weathering of the rock and sedimentation are decomposition processes. Through time, the minerals in the rocks soften due to pressure and heat. So, they crumble down and reduce in terms of size. Once they do, they become sand or part of the soil. So, the answer is A.
Answer:
5: 0.16
6: 50
Explanation:
Question 5:
We can use the equation density = mass/ volume.
We already have the mass (12g), but now we need to find the volume of the cylinder.
The equation for this is πr²h
So we know the radius is 2 and the height is 6.
π x (2)² x 6 = 24π = 75.398cm³
Now we can use the density equation above:
12/75.398 = 0.1592g/cm³ = 0.16g/cm³.
Question 6:
This time, we have to rearrange the equation density = mass/ volume to find the mass.
We know mass = density x volume.
From the question, the density is 2.5g/mL and the volume is 20mL.
Following the equation above, we do 2.5 x 20 to get 50g.
This reaction is most likely to fall under SN2 because the
thing called carbonication does not occur in SN1. The carbon forms a partial
bond with the nucleophile during the intermediate phase and the leaving group.
So for this question the reaction will fall under SN2.
Preparing 15 mg/gl working standard solution from a 20 mg/dl stock solution will require the application of the dilution principle.
Recalling the principle:
initial volume x initial molarity = final volume x final molarity
Since we were not given any volume to work with, we can as well just take an arbitrary volume to be prepared. Let's assume that the stock solution is 10 mL and we want to prepare 15 mg/gl from it:
Applying the dilution principle:
10 x 20 = final volume x 15
final volume = 200/15
= 13.33 mL
This means that in order to prepare 13.33 mL, 15 mg/l working standard solution from 10 ml, 20 mg/dl stock solution, 3.33 mL of the diluent must be added to the stock solution.
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