Each of the solutions was prepared with the same concentration so that the effect of concentration (0.05 M) would not be introdu
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This problem is providing us with the chemical equation depicting the production of ammonia from nitrogen and hydrogen at equilibrium and asks for the correct change when the concentration of nitrogen is increased. At the end, the answer is the forward reaction would increase to start reducing the concentration of N2.
<h3>Chemical equilibrium</h3>In chemistry, chemical reactions not always reach a 100-% conversion when reactants get in contact in order to carry out the chemical reaction. Thus, there is a point wherein the concentrations remain the same and is called equilibrium.
In such a way, for this problem, we have the following chemical reaction at equilibrium:
Now, according to the Le Ch.atelier's principle, an increase in the concentration of any species, shifts the equilibrium away from it, which means that if we increase the concentration of nitrogen, a reactant, the forward reaction will be favored.
Thereby, the correct answer is "the forward reaction would increase to start reducing the concentration of N2".
Learn more about chemical equilibrium: brainly.com/question/26453983
Hello!
To calculate the household concentration of NaOH we need to use the dilution formula, clearing for M2, as you can see in the equation below:
Now, we input the values from the data we have onto this equation. M1=19,1 M; V1=10 mL; V2=400 mL, and solve the equation to get the result:
So, the household concentration of NaOH will be 0,48 M
Have a nice day!
To calculate the household concentration of NaOH we need to use the dilution formula, clearing for M2, as you can see in the equation below:
Now, we input the values from the data we have onto this equation. M1=19,1 M; V1=10 mL; V2=400 mL, and solve the equation to get the result:
So, the household concentration of NaOH will be 0,48 M
Have a nice day!
Answer:
32
Step-by-step explanation:
There are two ways you can count the valence electrons.
A. From the Periodic Table
1 × P (Group 15) = 5
4 × O (Group 16) = 4 × 6 = 24
+3 e⁻ (for the charges) = <u> 3</u>
Total = 32
B. From the Lewis structure
In the <em>Lewis structure</em> (below), each line (bond) represents a pair of bonding electrons, and each dot represents an unbound electron (half a lone pair).
5 lines (bonds) = 5 × 2 = 10
3 single-bonded O atoms = 3 × 6 = 18
1 double-bonded O atom = <u> 4</u>
Total = 32
