You are balancing an ionic compound. Let's call it: AB.
1 answer:
Answer:-
Explanations:- It is given that the charge for A is +2 and the charge for B is -3. The over all compound is neutral means the over all charge is zero. For making the over all charge zero, we need 3 positive ions and 2 negative ions. This makes a +6 charge for A and -6 charge for B and the over all charge is zero.
Also, if we think about the criss cross then charge of A becomes the subscript of B and the charge of B becomes the subscript of A.
So, the formula of the ionic compound is . In this compound the ratio of A to B is 3:2.
The balanced equation could be shown as:
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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.