Answer:
thomson developed the chocolate chip method which was the identification of the electrons in the core of an atom. Rutherford discovered that the core was only positive and that the electrons were floating outside of the core.
Explanation:
Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
Answer:
A)
1. Reaction will shift rightwards towards the products.
2. It will turn green.
3. The solution will be cooler..
B) It will turn green.
Explanation:
Hello,
In this case, for the stated equilibrium:
In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:
A)
1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.
2. The formation of the green complex is favored, therefore, it will turn green.
3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.
B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.
Regards.
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.
Answer:
The correct answer is 160.37 KJ/mol.
Explanation:
To find the activation energy in the given case, there is a need to use the Arrhenius equation, which is,
k = Ae^-Ea/RT
k1 = Ae^-Ea/RT1 and k2 = Ae^-Ea/RT2
k2/k1 = e^-Ea/R (1/T2-1/T1)
ln(k2/k1) = Ea/R (1/T1-1/T2)
The values of rate constant k1 and k2 are 3.61 * 10^-15 s^-1 and 8.66 * 10^-7 s^-1.
The temperatures T1 and T2 are 298 K and 425 K respectively.
Now by filling the values we get:
ln (8.66*10^-7/3.61*10^-15) = Ea/R (1/298-1/425)
19.29 = Ea/R * 0.001
Ea = 160.37 KJ/mol
