A student takes 45.0 grams of N204 and combines it with 250mL of NaOH.
1. Name each compound.
2. State whether each compound bonds via ionic, covalent, or metallic
bonds.
3. Find the molarity of the final solution.
4. Identify whether the final solution is concentrated or dilute.
5. Identify the percent composition of oxygen in the final solution.
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The compounds involved in the reaction are nitrogen dioxide (NO2) and sodium nitrite (NaNO2).
Nitrogen dioxide (NO2) is a covalent compound, while sodium nitrite (NaNO2) is an ionic compound.
To find the molarity of the final solution, we need to first determine the number of moles of NaOH used. The molarity (M) of a solution is defined as moles of solute per liter of solution. The formula to calculate molarity is:
Molarity = (moles of solute) / (volume of solution in liters)
The volume of the solution is given in milliliters (mL), so we need to convert it to liters by dividing by 1000:
Volume of solution = 250 mL / 1000 mL/L = 0.25 L
The number of moles of NaOH used can be calculated using its concentration (assuming it's a standard solution), and the volume used:
moles of NaOH = concentration x volume of NaOH in liters
We don't have the concentration of NaOH, but we know that the reaction between N2O4 and NaOH is a neutralization reaction. This means that the number of moles of NaOH used is equal to the number of moles of H+ ions produced by the dissociation of N2O4. Since N2O4 is a covalent compound, it does not dissociate completely in water. However, it does react with water to form nitrous acid (HNO2):
N2O4 + H2O → 2HNO2
Each mole of N2O4 reacts with two moles of NaOH, and produces two moles of H+ ions. Therefore, the number of moles of NaOH used is equal to half the number of moles of N2O4 used:
moles of NaOH = 0.5 x (45.0 g / 92.01 g/mol) = 0.2457 mol
Now we can calculate the molarity of the final solution:
Molarity = (moles of NaOH) / (volume of solution in liters) = 0.2457 mol / 0.25 L = 0.983 M
Therefore, the molarity of the final solution is 0.983 M.
To determine whether the final solution is concentrated or dilute, we need to compare its molarity to the molarity of a typical solution of NaOH. A typical solution of NaOH has a molarity of about 1 M. Since the molarity of the final solution is less than 1 M, we can say that it is dilute.
To find the percent composition of oxygen in the final solution, we need to determine the number of moles of oxygen in the solution and the total number of moles of all the elements in the solution.
From the balanced equation, we know that each mole of N2O4 produces two moles of NO2, and each mole of NO2 contains one mole of oxygen:
N2O4 + 2NaOH → 2NO2 + 2H2O
This means that the number of moles of oxygen in the solution is twice the number of moles of N2O4 used:
moles of O = 2 x (45.0 g / 92.01 g/mol) = 0.978 mol
To find the total number of moles of all the elements in the solution, we need to add the number of moles of Na, N, and H:
moles of Na = 0.2457 mol
moles