<span>C2Br2
First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is
PV = nRT
where
P = pressure (1.10 atm = 111458 Pa)
V = volume (10.0 ml = 0.0000100 m^3)
n = number of moles
R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) )
T = Absolute temperature
Solving for n, we get
PV/(RT) = n
Now substituting our known values into the formula.
(111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol))
= (1.11458/2398.721652) mol
= 0.000464656 mol
Now let's calculate the empirical formula for this compound.
Atomic weight carbon = 12.0107
Atomic weight bromine = 79.904
Relative moles carbon = 13.068 / 12.0107 = 1.08802984
Relative moles bromine = 86.932 / 79.904 = 1.087955547
So the relative number of atoms of the two elements is
1.08802984 : 1.087955547
After dividing all numbers by the smallest, the ratio becomes
1.000068287 : 1
Which is close enough to 1:1 for me to consider the empirical formula to be CBr
Now calculate the molar mass of CBr
12.0107 + 79.904 = 91.9147
Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So
91.9147 g/mol * 0.000464656 mol = 0.042708701 g
0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is.
0.08541 / 0.0427087 = 1.99982673
1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.</span>
4.42
Because when you divide 129/29.20, you get a long string of numbers. 4.417808219178082...
So you round to the significant figure which in this case is 2 decimal places because 29.20 has 2 decimal places.
PS did you draw that car? Cuz im into drawing cars too.
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
Answer:
Because it went through a chemical change which changes its atomic form
ADD THEM all, and then divide by four. Thats what I would do!
