The molar mass of the gas is 77.20 gm/mole.
Explanation:
The data given is:
P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams
Using the ideal Gas Law will give the number of moles of the gas. The formula is
PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole
Also number of moles is not given so applying the formula
n= mass ÷ molar mass of one mole of the gas.
n = m ÷ x ( x molar mass) ( m mass given)
Now putting the values in Ideal Gas Law equation
PV = m ÷ x RT
3.29 × 4.60 = 37.96/x × 0.08206 × 375
15.134 = 1168.1241 ÷ x
15.134x = 1168.1241
x = 1168.1241 ÷ 15.13
x = 77.20 gm/mol
If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.
Convert mols to grams by multiplying grams of tin by the number of mols.
There are 119 grams per mol
119 x 11.8 = 1404 grams
Answer : The concentration of 
and 
are 
and 
respectively.
Solution : Given,
pH = 4.10
pH : pH is defined as the negative logarithm of hydronium ion concentration.
Formula used : ![pH=-log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D)
First we have to calculate the hydronium ion concentration by using pH formula.
![4.10=-log[H_3O^+]](https://tex.z-dn.net/?f=4.10%3D-log%5BH_3O%5E%2B%5D)
![[H_3O^+]=antilog(-4.10)](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3Dantilog%28-4.10%29)
![[H_3O^+]=7.94\times 10^{-5}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D7.94%5Ctimes%2010%5E%7B-5%7D)
Now we have to calculate the pOH.
As we know, 
Now we have to calculate the hydroxide ion concentration.
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![9.9=-log[OH^-]](https://tex.z-dn.net/?f=9.9%3D-log%5BOH%5E-%5D)
![[OH^-]=antilog(-9.9)](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dantilog%28-9.9%29)
![[OH^-]=1.258\times 10^{-10}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.258%5Ctimes%2010%5E%7B-10%7D)
Therefore, the concentration of and are and respectively.
<span>because p6 will be the group 8. You have to count the 2 electrons from the "s" block that are Group I and Group II
Group I s1
Group II s2
Group III s2 p1
Group IV s2 p2
Group V s2 p3
Group VI s2 p4
Group VII s2 p5
Group VIII s2 p6</span>
