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sukhopar [10]
4 years ago
6

What are the Group 1A and Group 7A elements examples of?

Chemistry
2 answers:
Galina-37 [17]4 years ago
8 0

Answer:

B, A, B, A are the answers for conexus

Explanation:

No, i'm not kidding you.

Scrat [10]4 years ago
6 0
Alkali metals = Group 1A
Halogens = Group 7A
You might be interested in
Which of these is not a commonly mined metallic substance?
Sliva [168]

Answer:

C) quartz

Explanation:

Quartz is a common mineral. Also, quartz isn't a metallic substance. I'm hoping that this helps :)

5 0
3 years ago
Which is the correct net ionic equation for the reaction of aqueous ammonia with nitric acid? Which is the correct net ionic equ
rewona [7]

Answer:

The correct answer is

e. NH3(aq) + H+(aq) --> NH4+(aq)

Explanation:

To solve this, we write out the indidual ionization reation for aqueous ammonia and nitric acid thus

For aqueous ammonia we have

NH₃(aq) + H₂O(l) ↔ NH₄⁺(aq) + OH⁻(aq)

Aqueous ammonia is a weak base and therefore undergoes partial ionization hence the reversible reaction sign

As the level of ionization will not be more than 5% OH⁻ cannot represebt the weak base

For nitric acid we have

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

a strong acid like nitric acid undergoes conplete ionization in the solution

The total equation is NH₃(aq) + HNO₃(aq) → NH₄NO₃(aq)

The sum of the ionic equation is

NH₃(aq) + H⁺(aq) + NO₃⁻(aq) → NH₄⁺(aq) + NO₃⁻(aq)

The ionic equation is

NH₃(aq) + H⁺(aq) →  NH₄⁺(aq)

5 0
4 years ago
Read 2 more answers
Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
Firlakuza [10]

Answer:

  • [Ag⁺] = 1.3 × 10⁻⁵M
  • s = 3.3 × 10⁻¹⁰ M
  • Because the common ion effect.

Explanation:

<u></u>

<u>1. Value of [Ag⁺]  in a saturated solution of AgCl in distilled water.</u>

The value of [Ag⁺]  in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                    0

C          -s                      +s                  +s

E         X - s                   s                     s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × s
  • Ksp = s²

By substitution:

  • 1.8 × 10⁻¹⁰ = s²
  • s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

  • [Ag⁺] = 1.3 × 10⁻⁵M ← answer

<u></u>

<u>2. Molar solubility of AgCl(s) in seawater</u>

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                  0.54

C          -s                      +s                  +s

E         X - s                   s                     s + 0.54

The new equation for the Ksp equation will be:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × ( s + 0.54)
  • Ksp = s² + 0.54s

By substitution:

  • 1.8 × 10⁻¹⁰ = s² + 0.54s
  • s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

     

     s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

<u>3. Why is AgCl(s) less soluble in seawater than in distilled water.</u>

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

3 0
3 years ago
Put the elements in order from easiest to hardest to remove an electron. Ba,Be,Ca,Sr.
laiz [17]

Answer:

Ba,Sr,Ca,Be

Explanation:

as atomic size increases ionisation potential decreases

8 0
3 years ago
Find the weighted average of these values. Col1 Value 5.00 6.00 7.00Col2 Weight 75.0 % 15.0 % 10.0 %
Nadya [2.5K]

Answer:

5.35

Explanation:

Value   5.00        6.00       7.00

Weight 75.0%     15.0%      10.0 %

We can determine the weighted average of these values using the following expression.

Weighted average = ∑ wi × xi

where,

w: relative weight

x: value

Weighted average = 5.00 × 0.750 + 6.00 × 0.150 + 7.00 × 0.100

Weighted average = 5.35

7 0
3 years ago
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