A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures wide and high. The maximum safe pressure inside the vessel has been measured to be . For a certain reaction the vessel may contain up to of boron trifluoride gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Be sure your answer has the correct number of significant digits.

The given question is incomplete, the complete question is:

A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 18.0cm wide and 21.6cm high. The maximum safe pressure inside the vessel has been measured to be8.50MPa For a certain reaction the vessel may contain up to0.362kg of carbon dioxide gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius.Be sure your answer has the correct number of significant digits.

The correct answer is 411 degree C.

Explanation:

The width of the vessel is 18.0 cm or 9.0 cm in radius and height of the vessel is 21.6 cm. In the given question, to find the volume of the cylinder, the formula to be used is pie r^2h

= 22/7 * (9)^2 * 21.6 cm^3

= 5498.7 cc

= 5.5 L

The value of 1MPa is 9.869 atm, therefore, the value of 8.5 MPa will be 83.89 atm.  

The maximum safe operating temperature can be calculated by using the ideal gas equation, that is, P = nRT/V or T = PV/nR

= 83.89 atm * 5.5 L / (0.362*10^3 g/44g/mol) * 0.082 L atmK^-1mol^-1

= 683.9 K

= 683.9 - 273 = 410.9 degree C = 411 degree C.  

Hence, the maximum safe operating temperature is 411 degree C.  

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'