Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
15 is the group that phosphorus is found in.
Answer : The final volume of gas will be, 26.3 mL
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,

where,
= initial pressure of gas = 0.974 atm
= final pressure of gas = 0.993 atm
= initial volume of gas = 27.5 mL
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:


Therefore, the final volume of gas will be, 26.3 mL
Answer: 1s^22s^22p^63s^23p^3
Explanation:
Assuming that orbital configuration is the same as electron configuration this is the answer.