Arrange the elements in order of increasing atomic radius. Use the periodic table to help you.
2 answers:
Answer: Order of atomic radius follows:
<u>Explanation:</u>
Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom.
As moving from top to bottom, there is an addition of shell around the nucleus and the outermost shell gets far away from the nucleus and hence, the distance between the nucleus and outermost shell increases. Thus, increasing the atomic radii of the atom.
As moving from left to right in a period, more and more electrons get added up in the same shell and the attraction between the last electron and nucleus increases, which results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom on moving towards right of the periodic table.
For the given elements:
Magnesium is present in Group 2, Period 3 of the periodic table.
Barium is present in Group 2, Period 6 of the periodic table.
Fluorine is present in Group 17, Period 2 of the periodic table.
Sulfur is present in Group 16, Period 3 of the periodic table.
The increasing order of atomic radius follows:
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The question requires us to calculate the mass of NaNO3 necessary to prepare 506 mL of a 10.5 M solution of this salt.
To answer this question, we'll need to go through the following steps:
1) Calculate the molar mass of NaNO3, as it will be necessary during the question;
2) Calculate the number of moles necessary to prepare 506 mL of the 10.5 M solution;
3) Convert the number of moles calculated into the mass of NaNO3, using the molar mass of this salt.
Next, we'll follow the steps:
1) To calculate the molar mass of NaNO3, we need to consider the atomic masses of Na, N and O and the number of atoms of each of these elements. The atomic masses are 22.99 u, 14.01 u and 15.99 u for Na, N and O, respectively. With this information, we can calculate the molar mass:
molar mass of NaNO3 = (1 * 22.99) + (1 * 14.01) + (3 * 15.99) = 84.97 g/mol
2) Next, we use the molar concentration provided (10.5 M or 10.5 mol/L) and the volume required (506 mL = 0.506 L) to obtain the number of moles necessary to prepare this solution:
<em>1 L of solution --------------- 10.5 mol of NaNO3</em>
<em>0.506 L of solution -------- x</em>
Solving for x, we have:
Therefore, the number of moles of NaNO3 necessary to prepare the solution is 5.31 moles.
3) At last, we use the molar mass obtained in step 1 (84.97 g/mol) to calculate the mass that corresponds to 5.31 moles of NaNO3:
<em>1 mol of NaNO3 ---------------- 84.97 g</em>
<em>5.31 mol of NaNO3 ----------- y</em>
Solving for y, we have:
Therefore, the mass of NaNO3 necessary to prepare the solution given by the question is 451.2 g.