Lactose, a sugar in milk, is composed of one glucose molecule joined by a glycosidic linkage to one galactose molecule. how is l

Answer:

45727g

Explanation:

The overall ionic reaction can be described as follows:

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

The amount of wastewater processed each day is specified as 60m^3/h, with the concentration of chromium in the wastewater recorded at 4.0 mg/L, and the allowable discharge concentration is set at 0.1 mg/L.

Step one: Convert m^3/h to L/h. Therefore, 60 m^3/h × 1000 dm^3 = 60000 L/h.

Step two: Calculate the amount of chromium consumed.

The amount of chromium utilized = { 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Calculate the masses of Cr(OH)3 and Fe(OH)3.

The moles of chromium = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And, the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Consequently, the total equals 463.5 g + 1441.8 g = 1905.3 g.

Step four: Calculate the amount of particulate matter generated each day.

The total particulate matter produced daily = 24 × 1905.3 = 45727g.

CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O

m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}

0.2845/{44.01x}=0.1451/{9.01y}

x/y=0.4=2:5

The empirical formula is C₂H₅.

I believe the answer is D, though I'm not entirely certain unfortunately.

The quantity of fluorine in moles is calculated as 71/19 = 3.74
We also know that at standard temperature and pressure (273 K and 101.3 kPa), one mole of gas occupies 22.4 liters
So, the volume for 3.74 moles at S.T.P is: 3.74 x 22.4
This results in a volume of 83.776 L, which is equivalent to 83,776 mL

Next, applying Boyle's law, which states that for a fixed amount of gas,
PV = constant

We set up the equation P x 6843 = 101.3 x 83776
Solving for P gives us 1,240 kPa

Based on the titration results, the adult would need to consume 85.7 mL to meet the recommended daily intake of 60 mg of Vitamin C. This is calculated from the average total volume of DCPIP used during trials.