The formation of ethyl alcohol (C2H5OH) by the fermentation of glucose (C6H12O6) may be represent by the following: C6H12O6 --> 2 C2H5OH 2 CO2 If a particular glucose fermentation process is 70.0% efficient, how many grams of glucose would be required for the production of 51.0 g of ethyl alcohol (C2H5OH)

142.5 g

Explanation:

According to the chemical reaction:

C₆H₁₂O₆ --> 2 C₂H₅OH + 2 CO₂

1 mol of glucose (C₆H₁₂O₆) forms 2 moles of ethyl alcohol (C₂H₅OH) and 2 moles of carbon dioxide (CO₂).

We first convert the moles to grams by using the molecular weight (Mw) of each compound:

Mw (C₆H₁₂O₆) = (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6)= 180 g/mol

1 mol C₆H₁₂O₆ = 180 g/mol x 1 mol = 180 g

Mw(C₂H₅OH) = (12 g/mol x 2) + (1 g/mol x 5) + 16 g/mol + 1 g/mol= 46 g/mol

2 mol C₂H₅OH = 2 mol x 46 g/mol = 92 g

Thus, when the process is 100% efficient, 180 grams of glucose produce 92 grams of ethyl alcohol. To form 51.0 grams of ethyl alcohol, we will need:

51.0 g C₂H₅OH x (180 g C₆H₁₂O₆/92 g C₂H₅OH) = 99.8 g C₆H₁₂O₆

As the process has a lower efficiency (70.0%), we will need more glucose to obtain the required yield. So, we divide the mass of glucose required for a process 100% efficient by the actual efficiency:

mass of glucose required = 99.8 g C₆H₁₂O₆/(70%) = 99.8 g C₆H₁₂O₆ x 100/70 = 142.5 g

Therefore, it would be required 142.5 grams of glucose to obtain 51.0 grams of ethyl alcohol.

Answer:

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the periodic table)

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.