What is the temperature (°c) of 1.75 g of o2 gas occupying 4.60 l at 1.00 atm?

Given- m= 1.75 g  

Now moles of O2 gas= Mass/ Molar mass  

n = 1.75/32  

V= 2L  

P= 1atm  

R=0.0821Constant  

As PV=nRT  

=> T=PV/nR  

= 1* 2/ (1.75/32)*0.0821 K  

= 2*32/1.75* 0.0821 K  

= 445.45 K  

= (445.45- 273.15) degrees celsius  

= 172.30 degrees celsius ANSWER

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.