We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
Q1. Chemical, Physical, Physical, Physical
(l am not 100% sure about the 4th answer)
Q2. All of the above
It's useful because it highlights structures in biological tissue.
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, 
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in 
of ethane gas
And, as we know that
1 mole of ethane molecule contains 
molecules of ethane
2.869 moles of ethane molecule contains 
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
<h3><u>Answer;</u></h3>
<u>= 5 M or 5 moles/liter</u>
<h3><u>Explanation;</u></h3>
At point E, 90 g of substances X are dissolved in 100 g of the solvent.
100g of the solvent is equal to 100 ml
Molarity is the number of moles of a substance in one liter of a solvent.
90 g of X are in 100 ml
But; the RFM of X = 180 g/l
Therefore; the moles of X in 90 g = 90/180
= 0.5 moles
Therefore;
0.5 moles of X are contained in 100 ml of the solvent;
Thus, molarity = 0.5 × 1000/100
=<u> 5 M or 5 moles/liter</u>
