First convert 12.0g of Na to moles using the grams to moles conversation and you get about .5219 moles (I didn't use significant figures). Divide that number by 3 because that is the coefficient of Na and you will get about .17398 moles, which is how many moles are in 1. Since Al has a coefficient of 1, .17398 woild be your final answer
Answer:
B) 0.59 M NaCl.
Explanation:
- It is known that the no. of millimoles of NaCl before dilution = the no. of millimoles of NaCl after dilution.
∵ (MV) before dilution = (MV) after dilution.
<em>∴ M after dilution = (MV) before dilution / V after dilution </em>= (3.2 M)(25.0 mL)/(135.0 mL) = <em>0.5926 M ≅ 0.59 M.</em>
Answer:
B. ADDITION OF TWO GROUPS ACROSS A DOUBLE BOND
Explanation:
Addition reaction of alkenes involves the conversion of the double bond in alkenes Inyo single bonds by the addition of two groups of atoms or radicals.
During this addition reaction, two substances, an unsaturated compound(e.g. ethane) and an attacking reagent (hydrogen, halogens, hydrogen halides, chlorine and bromine water) combines to form a single new compound without forming any other products. So a saturated product or one in which is an increase in degree of saturation is formed.
Answer:
5.01%
Explanation:
Density of vinegar = mass/volume
Mass of 10.00 mL = density x volume
= 1.006 x 10 = 10.06 g
From the equation of reaction:
1 mole pf CH3COOH requires 1 mole of NaOH for neutralization.
mole of NaOH = molarity x volume
= 0.5062 x 0.01658
= 0.008392796 mole
0.008392796 mole of NaOH will therefore require 0.008392796 mole of CH3COOH.
mass of CH3COOH = mole x molar mass
= 0.008392796 x 60.052
= 0.504 g
Percentage by mass of acetic acid in the vinegar = 0.504/10.06 x 100%
= 5.01%
The percent by mass of acetic acid in the vinegar is 5.01%
