It is so simple when you have [OH-] so you will use it to get POH from this formula:
POH = -㏒[OH]
= -㏒ 2.5 x 10^-4
= 3.6
when PH + POH = 14
∴ PH = 14 - POH
= 14 - 3.6 = 10.4
If water was nonpolar, life as it exists now would be impossible. Water is the means by which ionic substances ( Fe^3+, Na^+, Ca^2+, K^+, etc) and polar molecules are distributed around and in the cells of organisms.
<span>Water is a major contributor to erosion of mountains because of its polar nature. It is able to dissolve almost all minerals because of its attraction for the positive and negative parts of the minerals that it encounters. </span>
<span>If it was nonpolar, it would not exist in the solid and liquid state on earth. It is the attraction of the oppositely charged ends of different water molecules that makes it assume the liquid and solid state under the temperature ranges found on earth.</span>
Answer:
The molarity of the acetic acid in this vinegar is 0.853 M
Explanation:
Step 1: Data given
Volume of vinegar sample = 10.00 mL
Concentration of NaOH = 0.5052 M
16.88 mL are required to neutralize the acetic acid
Step 2: The reaction
HC2H3O2(aq) + NaOH(aq) ⇔ NaC2H3O2(aq) + H2O(l)
Step 3: Calculate molarity of cetic acid (HC2H3O2)
C1*V1 = C2*V2
⇒with C1 = the molarity ofHC2H3O2= TO BE DETERMINED
⇒with V1 = the volume of HC2H3O2 = 10.00 mL = 0.01 L
⇒with C2 = the molarity of NaOH = 0.5052 M
⇒with V2 = the volume of NaOH = 16.88 mL = 0.01688L
C1 = (C2*V2)V1
C1 = (0.5052 * 0.01688 ) / 0.01
C1 = 0.853 M
The molarity of the acetic acid in this vinegar is 0.853 M
Physical changes are used to separate mixtures into their component compounds, but can not usually be used to separate compounds into chemical elements or simpler compounds.Physical changes occur when objects or substances undergo achange that does not change their chemical composition.
A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.
given that :
8 ml of a 5 % solution mix with 10 ml . that means the 80 mL of 5 % solution is diluted with water of 10 mL
therefore, 80 × 5 = 10 × x %
x % = 40 %
Therefore, the final percentage strength of the solution is 40 %
Thus, A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.
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