X will be in group 5, since if you exchange the valencies of Na with any element on group 5, you will get Na3X
The model would look something like the image below.
There would be a <em>central nucleus</em> containing <em>20 protons</em> and <em>20 neutrons</em>.
Surrounding the nucleus would be four concentric rings (energy levels) containing <em>20 electron</em>s.
Going out from the nucleus, the number of electrons in each ring would
be <em>2, 8, 8, 2</em>.
Answer:
0.3229 M HBr(aq)
0.08436M H₂SO₄(aq)
Explanation:
<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>
<em />
Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).
NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)
When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>
<em />
Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).
2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)
When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

Answer:
Upper F subscript 2 (g) plus upper C a (s) right arrow with delta above upper C a upper F subscript 2 (s).
Explanation:
This is a chemical reaction problem.
In expressing any chemical reaction, we need to understand that there are reactants and products.
- The reactants are the species on the left hand side that are combining.
- The products are the species on the right hand side that are formed.
- Every chemical reaction is obeys the law of conservation of matter i.e equal number of matter on both sides.
Using the statement of this problem, we can deduce that;
Reactants are Fluorine gas and Calcium metal
Product is Calcium Fluoride
Note: A metal is a solid(s) and powder is a solid(s). A gas is denoted as (g). They depict the state of the species reacting.
F₂
+ Ca
→ CaF₂
We can see that equal number of atoms are on both sides of the expression.