Mass, if you know what element you are working with.
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
Answer:
C: 4
H: 1
Mg: 2
O : 6
Explanation:
You can quickly find out the number of valence electrons by looking at where the element is on the periodic table and referring to the table that is attached.
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
