Is there options? If not, then your answer is going to be cells!
Answer:
see explanation and punch in the numbers yourself ( will be better for your test)
Explanation:
If you are given atoms you need to divide by Avogadro's number 6.022x10^23
then you will have moles of sulfur-- once you have moles multiply by the molar mass of sulfur to go from moles to grams
mm of sulfur is 32 g/mol
Answer:
20 hydrogen atoms
Explanation:
Let us carefully observe the words of the question. We were told that the compound is a hydrocarbon. This implies that it is composed of only hydrogen and carbon.
Secondly, we are informed that the compounds is composed of only C-C single covalent bonds. This means that the compound is an alkane.
Then we are told that it contains nine carbon atoms and is a linear molecule. Hence, the compound must be n-nonane.
Recall that for alkanes, the general molecular formula is CnH2n+2 Substituting n=9 for nonane we have; C9H(2*9)+2 which is C9H20.
Thus there are 20 hydrogen atoms. The structure of the compound is attached to this answer
Answer:
17.27 atm
Explanation:
Use the ideal gas law or PV = nRT
We are solving for pressure here so lets isolate for P before we plug in values:
![P = \frac{nRT}{V}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BnRT%7D%7BV%7D)
So first to get n or the number of moles we need to convert the grams of N2O to moles of N2O. We can do this by multiplying by the inverse of the molar mass like so:
![\frac{519.93g(N2O)}{1} \frac{mol}{44.013g(N2O)}](https://tex.z-dn.net/?f=%5Cfrac%7B519.93g%28N2O%29%7D%7B1%7D%20%5Cfrac%7Bmol%7D%7B44.013g%28N2O%29%7D)
Our grams of N2O would cancel and give us 11.813 mol of N2O
Now all thats left is to plug in and solve with the correct value for R which in this case for all of our units to cancel is 0.08206
![P = \frac{(11.813)(0.08206)(125.63)}{7.05}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%2811.813%29%280.08206%29%28125.63%29%7D%7B7.05%7D)
P = 17.27 atm
(I would double check the calculator work if it is for correctness just be sure)
I would go with a iconic Bond.