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user100 [1]
2 years ago
6

What does B represent?

Chemistry
1 answer:
Mashutka [201]2 years ago
8 0

Answer: Botanical Acid

Explanation:

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Need helpppp please
Vladimir79 [104]

Answer:

message me so i can help you because i cant see the words

Explanation:

6 0
2 years ago
A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
vesna_86 [32]

Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 0.375 M

V_1 = volume of  NaOH solution = 62.5 ml

Putting in the values we get:

2\times M_1\times 50.0=1\times 0.375\times 62.5

M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

6 0
2 years ago
In the background, you were given the reactions for the fermentation of glucose to ethanol in the production of wine from grapes
ikadub [295]

Answer:

Kombucha is a Symbiotic culture of bacteria and yeast (SCOBY), similar to mother of vinegar, containing one or more specie each of bacteria and yeast which is added to sweet tea, which jump start two processes take place. First, alcoholic fermentation during which the yeast convert sugars to alcohol under controlled conditions. After that, the bacteria converts most of this alcohol into acetic, gluconic and glucoronic and other organic acids. during both of this processes, the yeast and bacteria, feast and multiply, yielding an end beverage that is rich in variety of micro organisms and healthy acids. Source: Health-Ade.com/blog/blog/what-is-fermentation

the products formed is more rich than those of the fermentation of grapes because it is a symbiotic fermentation of yeast and bacteria

Explanation:

Kombucha is fermented slightly alcoholic, lightly effervescent sweetend black or green tea, commonly consumed for its health benefits sometimes the beverage is called kombucha tea to distinguish it from the cultures of bacteria and yeast. Kombucha is thought to have originated in manchuria where the drink is traditionally consumed or in Russia and Eastern Europe.

Source: "A mug Of Kombucha for your health?"     .<em>mayo clinic. </em>Retrieved 2018-09-01

7 0
2 years ago
Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the d
Elina [12.6K]

Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

8 0
3 years ago
An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm
frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

 xA= 0.34, yA= 0.55

ii)      0.50 = 0.55 nv + 0.34 nL

     Therefore, nV =    0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii)  ρA= 0.791 g/cm3 and,  ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

3 0
3 years ago
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