Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.
Solution:- The balanced equation for the combustion of acetylene is:

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

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The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.
From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

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Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.
So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

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(a) Iron (iii) sulphate:
From the periodic table:
mass of iron = 55.845 grams
mass of sulphur = 32.065 grams
mass of oxygen = 16 grams
Iron (iii) sulphate has the formula: Fe2(SO4)3
molar mass = 2(55.845) + 3(32.065) + 3(4)(16) = 399.885 grams
(b) Sodium hydroxide:
From the periodic table:
mass of sodium = 22.989 grams
mass of oxygen = 16 grams
mass of hydrogen = 1 gram
Sodium hydroxide has the formula: NaOH
molar mass = 22.989 + 16 + 1 = 39.989 grams
(c) Barium carbonate
From the periodic table:
mass of barium = 137.327 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Barium carbonate has the formula: BaCO3
molar mass = 137.327 + 12 + 3(16) = 197.327 grams
(d) ammonium nitrate:
From the periodic table:
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
Ammonium nitrate has the formula: NH4NO3
molar mass = 14 + 4(1) + 14 + 3(16) = 80 grams
(e) Lead (iv) oxide
From the periodic table:
mass of lead = 207.2 grams
mass of oxygen = 16 grams
Lead (iv) oxide has the formula: PbO2
molar mass = 207.2 + 2(16) = 239.2 grams
From the above calculations, we can see that:
Iron (iii) sulphate has the greatest mass.
Answer:
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
Consider the example of sodium and potassium.
Sodium is present above the potassium with in same group i.e, group one.
The atomic number of sodium is 11 and potassium 19.
So potassium will have larger atomic radius as compared to sodium.
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
I don't see the options for an answer, so here is a list of all of the transition metals lol
- <em>Scandium</em>
- <em>Titanium</em>
- <em>Vanadium</em>
- <em>Chromium</em>
- <em>Manganese</em>
- <em>Iron</em>
- <em>Cobalt</em>
- <em>Nickel</em>
- <em>Copper</em>
- <em>Zinc</em>
- <em>Yttrium</em>
- <em>Zirconium</em>
- <em>Niobium</em>
- <em>Molybdenum</em>
- <em>Technetium</em>
- <em>Ruthenium</em>
- <em>Rhodium</em>
- <em>Palladium</em>
- <em>Silver</em>
- <em>Cadmium</em>
- <em>Lanthanum</em>
- <em>Hafnium</em>
- <em>Tantalum</em>
- <em>Tungsten</em>
- <em>Rhenium</em>
- <em>Osmium</em>
- <em>Iridium</em>
- <em>Platinum</em>
- <em>Gold</em>
- <em>Mercury</em>
- <em>Actinium</em>
- <em>Rutherfordium</em>
- <em>Dubnium</em>
- <em>Seaborgium</em>
- <em>Bohrium</em>
- <em>Hassium</em>
- <em>Meitnerium</em>
- <em>Darmstadtium</em>
- <em>Roentgenium</em>
- <em>Copernicium p</em>
The correct answer is option B. X-rays are the waves found in the electromagnetic spectrum at a wavelength of 1x10^-11 - 1x10^-8 m and frequencies of 3x10^16 to 3x10^19 Hz. These waves penetrate easily through certain materials which made them usefule in the field of medicine.