Note that it says oxygen "gas"
So you need the atomic mass of oxygen gas
Look at your periodic table, you'll see 15.9994 under oxygen
Oxygen gas has a formula of O2 therefore,
(15.9994) times 2= Oxygen gas atomic mass=31.9988
Mol= Mass/Atomic Mass
=62.3 g/ 31.9988 g/mol = 1.95 mol
now look at the ratio of C2H6 and O2, notice there is an invisible number beside each of them, at that "invisible number" is =1
1 C2H6 + 1 O2 -> products
this means that for 1 mol of C2H6, 1 mol of O2 has to react with it
Thus as we have 1.95 moles of O2, we need 1.95 moles of C2H6
Answer:
Se =[Ar] 3d¹⁰ 4s² 4p⁴
Explanation:
The noble gas notation is used for the shortest electronic configuration of other periodic table elements.
For example:
The atomic number of Argon is 18, and its electronic configuration is,
Ar₁₈ = 1s² 2s² 2p⁶ 3s² 3p⁶
The atomic number of selenium is 34, its electronic configuration is,
Se₃₄ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴
By using the noble gas notation, electronic configuration of selenium can be written is shortest form.
Se =[Ar] 3d¹⁰ 4s² 4p⁴
This electronic configuration is also called abbreviated electronic configuration.
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.