Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
So,
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
Answer:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
760.0 mmHg(V)=600.0mmHg(240.0mL)
600(240)/760=189mL
<h3>
Answer:</h3>
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
<h3>
Explanation:</h3>
Concept tested: Balancing of chemical equations
- A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
- Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
- In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
- Therefore; the balanced chemical equation for the reaction is;
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
The energy at n level of hydrogen atom energy level =13.6/n^2
substiture the respective n values in the equation above and find the difference in the energy levels
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