Answer:
atomic number- 42
chemical number- 96
number of protons-18
atomic mass-96
number of eletrons-42
number of valence eletrons- 1, 2, 13, or 18. (sorry i forgot)
nummber of shells in n=3 shell-18
number of neutrons-54
Explanation:
well i just know the periodic table by heart and thats how i got the answer XD
Answer:
- <em><u>The third chice: 0.0750 M Na₂SO₄</u></em>
Explanation:
Assume 100% ionization:
<em><u>1) 0.100 M solution K₂SO₄</u></em>
- K₂SO4 (aq) → 2K⁺ (aq) + SO₄²⁻ (aq)
- Mole ratios: 1 mol K₂SO4 : 2 mol K⁺ + 1 mol SO₄²⁻ (aq) : 3 mol ions. This is 1 : 3
- At constant volume, the mole ratios are equal to the concentration ratios (M).
- 1 M K₂SO₄: 3 M ions = 0.100 M K₂SO₄ / x ⇒ x = 0.300 M ions
This means, that you have to find which of the choices is a solution that contains the same 0.300 M ion concentration.
<u>2) 0.0800 M Na₂CO₃</u>
- Na₂CO₃ (aq) → 2 Na⁺ + CO₃⁻
- 1 M Na₂CO₃ / 3 M ions = 0.0800M / x ⇒ x = 0.0267 M ions
This is not equal to 0.300 M, so this solution would not contain the same total concentration as a 0.100 M solution of K₂SO₄, and is not the right answer.
<u>3) 0.100 M NaCl </u>
- 1 M NaCl / 2 M ions = 0.100 M NaCl / x ⇒ x = 0.200 M ions
This is not equal to 0.300 M ion, so not a correct option.
<u>4) 0.0750 M Na₃PO₄</u>
- 1 M Na₃PO₄ / 4 M ions = 0.0750 M Na₃PO₄ / x ⇒ x = 0.300 M ions
Hence, this ion concentration is equal to the ion concentration of a 0.100 M solution of K₂SO₄, and is the correct choice.
<u>5) 0.0500 M NaOH </u>
- 1 M NaOH / 2 mol ions = 0.0500 M NaOH / x ⇒ x = 0.100 M ions
Not equal to 0.300 M, so wrong choice.
Darwin used finches in his theory of evolution.
Balance Chemical equation is as follow,
<span> 3 H</span>₂ <span>(g) + N</span>₂ <span>(g) </span>→<span> 2 NH</span>₃ <span>(g)
According to balanced equation, 3 Molecules (3 moles) of Hydrogen reacts with 1 Molecule of N</span>₂ to produce 2 moles (2 Molecules) of NH₃.
Result:
2 Molecules of Ammonia are produced by reacting 3 molecules of Hydrogen and 1 molecule of Nitrogen.
Answer:
Average atomic mass of carbon = 12.01 amu.
Explanation:
Given data:
Abundance of C¹² = 98.89%
Abundance of C¹³ = 1.11%
Atomic mass of C¹² = 12.000 amu
Atomic mass of C¹³ = 13.003 amu
Average atomic mass = ?
Solution:
Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass of carbon = (12.000×98.89)+(13.003×1.11) /100
Average atomic mass of carbon= 1186.68 + 14.43333 / 100
Average atomic mass of carbon = 1201.11333 / 100
Average atomic mass of carbon = 12.01 amu.
