Most wine is prepared by the fermentation of the glucose in grape juice by yeast: c6h12o6(aq) --> 2c2h5oh(aq) + 2co2(g) how many grams of glucose should there be in grape juice to produce 725 mls of wine that is 11.0% ethyl alcohol, c2h5oh (d=0.789 g/cm3), by volume?

123.41 g

Explanation:

Given that the ethyl alcohol produced is 11.0 % by volume.

It means that 1000 mL contains 110 mL of ethyl alcohol

Given that the volume is:- 725 mL

So,

Volume of ethyl alcohol = = 79.75 mL

Given that:- Density = 0.789 g/cm³ = 0.789 g/mL

So, Mass = Density*Volume = = 62.92 g

Calculation of the moles of ethyl alcohol as:-

Molar mass of ethyl alcohol = 46.07 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the reaction:-

2 moles of ethyl alcohol is produced when 1 mole of glucose reacts

Also,

1.37 moles of ethyl alcohol is produced when mole of glucose reacts

Moles of glucose = 0.685 Moles

Molar mass of glucose = 180.156 g/mol

Mass = Moles*Molar mass = = 123.41 g

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j