3. What is the atomic mass of phosphorous if phosphorous-29 has a percent abundance of 35.5%, phosphorous-30 has a percent abund

The answer is.997 atm. 1. Identify the combined gas law equation... (P1V1/T1 = P2V2/T2) 2. Gather our values... P1=.982 atm P2=? (this is what we want to find) V1= 2 L V2= 1.8 L T1= 22 C = 295 K T2= -3 C = 270 K - Note: Always convert to Kelvin. To do this, add 273 to the Celsius value. 3. Rearranging the formula to suit this problem... (P2=P1V1T2/V2T1) 4. Insert our values... P2=.982 atm x 2 L x 270 K / 1.8 L x 295 K 5. Calculate and the result should yield....997 atm - If you require further clarification or assistance, feel free to reach out and I'll be happy to help!

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.

Response: 1-methoxy-2,4-dinitrobenzene

Rationale:

The nitro groups exhibit a strong electron-withdrawing effect, facilitating nucleophilic substitution reactions where a substituent is replaced by a robust nucleophile like the methoxy group. The reaction's mechanism is illustrated below. The electron-withdrawing nature of the nitro group aids in the formation of the intermediate during the reaction as depicted.