Answer:
The partial pressure of the other gases is 0.009 atm
Explanation:
Step 1: Data given
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.
The atmospheric pressure = 0.90 atm
Step 2: Calculate mol fraction
If wehave 100 moles of air, 78 moles will be nitrogen,
21 moles will be oxygen, and 1 mol will be other gases.
Mol fraction = 1/100 = 0.01
Step 3: Calculate the partial pressure of the other gases
Pgas = Xgas * Ptotal
⇒ Pgas = the partial pressure = ?
⇒ Xgas = the mol fraction of the gas = 0.01
⇒Ptotal = the total pressure of the pressure = 0.90 atm
Pgas = 0.01 * 0.90 atm
Pgas = 0.009 atm
The partial pressure of the other gases is 0.009 atm
Yes thats true! You always have to think about the question or project before you start a science experiment! :)
Answer:
a) cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)
b) 0.068 V.
Explanation:
A) Cu2+ + 2e- euilibrium cu (s)
Hg2Cl2 + 2e- equilibrium 2Hg (l) + 1cl-
Cell Reaction: cu2+ + 1Hg (l) 1Cl- equilibrium cu (s) + Hg2Cl2 (s)
B) To calculate the cell voltage
E = E_o Cu2+/Cu - (0.05916 V / 2) log 1/Cu2+
putting values we get
= 0.339V + (90.05916V/2)log(0.100) = 0.309V
E_cell = E Cu2+/Cu - E SCE = 0.309 V - 0.241 V = 0.068V.
a) The E might belong to group 13.
As the formula of a chemical compound is derived by the cross multiplication of the valency of the atoms. As formula of the given oxide is and valency of O atom is -2, therefore valency of element E must be +3 in order to obtain E2O3.
Also, in EF3, the valency of E will be +3 because there are three atoms of fluorine who has an individual valency of -1. Thus, e will have the valency of +3.
The Group 13 is the boron group which has the following elements:
- Boron
- Aluminium
- Gallium
- Indium
- Thallium
All these elements have the valency of +3.
To know more about Valency, refer to this link:
brainly.com/question/12717954
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