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
<span>All Right Cesium is an Alkali Metal so it Belongs to the first Column of the Periodic table. Every Element in that Column Has a Charge of 1+ as a cation. so at the answer your question, a single Cesium atom loses 1 electron when it became cation. This cation look like this:
cs=55 electrons
cs^+= 54 Electrons
As you Can see it only loses one</span>
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Please refer to this. Here is the oxidation number of nitrogen in aluminum nitride, aln. <span>Since AlN has 0 charge and Al has +3 charge; charge for N = 0 - (+3) = -3. Hope this is the answer that you are looking for. Have a great day ahead!</span>