Answer:
Molar Concentration = 
= 
= 13.33
No. of H+ ions present = 13.33
pH value = - log[13.33]
= -1.12
Explanation:
The equivalence point, or stoichiometric point, of a substance response is the point at which synthetically identical amounts of reactants have been blended. As such, the moles of corrosive are equal to the moles of base, as per the condition (this doesn't really infer a 1:1 molar proportion of acid:base, simply that the proportion is equivalent to in the condition). It tends to be found by methods for a marker, for instance phenolphthalein or methyl orange. The endpoint (identified with, however not equivalent to the equivalence point) alludes to the point at which the marker changes shading in a colorimetric titration.
First let's convert the minutes to hours (multiply minutes by 60 to get hours):
30 × 60 = 1,800 drops/hour
Now that you know how much drops there are per hour, you can multiply this answer by 3 to work out how many drops there are in 3 hours:
1,800 × 3 = 5,400 drops
We know that 5 drops is equal to 1 ml, so we can divide 5,400 by 5 to work out the amount of ml:
5,400 ÷ 5 = 1,080 ml
Therefore, your final answer is 1,080 millilitres (ml)
Is the water line of boat when someone boards or leaves it.
Answer:
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O.
Explanation:
- This is an acid-base reaction which produces salt and water according to the balanced equation:
<em>2KOH + H₂SO₄ → K₂SO₄ + 2H₂O.</em>
<em></em>
it is clear that 2.0 moles of KOH react with 1.0 mole of H₂SO₄ to produce 1.0 mole of K₂SO₄ and 2.0 moles of H₂O
To prepare 350 mL of 0.100 M solution from a 1.50 M
solution, we simply have to use the formula:
M1 V1 = M2 V2
So from the formula, we will know how much volume of the
1.50 M we actually need.
1.50 M * V1 = 0.100 M * 350 mL
V1 = 23.33 mL
So we need 23.33 mL of the 1.50 M solution. We dilute it
with water to a volume of 350 mL. So water needed is:
350 mL – 23.33 mL = 326.67 mL water
Steps:
1. Take 23.33 mL of 1.50 M solution
<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M
solution</span>
