When NaCH3Coo mixed with HCl we will get NaCl and CH3CooH as shown in the following balanced equation:
NaCH3Coo + HCl → NaCl + CH3CooH
so from this equation, we can conclude that is no precipitate because all we get is the acetic acid which found in vinegar and the NaCl which is very soluble so we don't have any precipitate.
so, your answer is no precipitate, no reaction
The answer: is yes, It is a buffer solution.
first, we need to get moles of sodium hydroxide and propanoic acid:
moles NaOH = molarity * volume
= 0.5M * 0.1 L = 0.05 moles
moles propanoic acid = molarity * volume
= 0.75 M * 0.1 L = 0.075 moles
[NaOH] at equilibrium = 0.05 m
[propanoic acid ] at equilibrium = 0.075 - 0.05 = 0.025 m
when Pka for propanoic acid (given) = 4.89
so by substitution:
∴PH = Pka + ㏒[NaOH]/[propanoic acid ]
∴ PH = 4.89 + ㏒ 0.05 / 0.025
= 5.19
Answer:
molecular weight of H2O2 or grams. This compound is also known as Hydrogen Peroxide. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles H2O2, or 34.01468 grams.
1 grams H2O2 is equal to 0.029399071224542 mole.
1 grams H2O2 to mol = 0.0294 mol
10 grams H2O2 to mol = 0.29399 mol
20 grams H2O2 to mol = 0.58798 mol
30 grams H2O2 to mol = 0.88197 mol
40 grams H2O2 to mol = 1.17596 mol
50 grams H2O2 to mol = 1.46995 mol
100 grams H2O2 to mol = 2.93991 mol
200 grams H2O2 to mol = 5.87981 mol
The graph is not given in the question, so, the required graph is attached below:
Answer:
According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.
The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.
When many atoms share electrons together.
