Answer:
Average density for method A = 2.4 g/cm³
Average density for method B = 2.605 g/cm³
Explanation:
In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):
Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6
Average = 9.6/4 = 2.4 g/cm³
Σ = 2.603 + 2.601 + 2.605 + 2.611 = 10.420
Average = 10.420/4 = 2.605 g/cm³
Since you have not included the chemical reaction I will explain you in detail.
1) To determine the limiting agent you need two things:
- the balanced chemical equation
- the amount of every reactant involved as per the chemical equation
2) The work is:
- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.
- determine the number of moles of each reactant with this formula:
number of moles = (mass in grams) / (molar mass)
- set the proportion with the two ratios (theoretical moles and actual moles)
- compare which reactant is below than the stated by the theoretical ratio.
3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:
i) Chemical equation: H₂ + O₂ → H₂O
ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O
iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂
iv) Covert 100 g of H₂ into number of moles
n = 100g / 2g/mol = 50 mol of H₂
v) Convert 100 g of O₂ to moles:
n = 100 g / 32 g/mol = 3.125 mol
vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂
vii) Compare the two ratios:
2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂
Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.
Therefore, the O₂ is the limiting reactant in this example.
Answer:
0.034M HCl is the concentration of the diluted solution
Explanation:
You take, initially, 25.00mL of the 0.136M HCl. Then, you dilute the solution to 100.00mL. The solution is diluted:
100.00mL / 25.00mL = 4. The solution was diluted 4 times.
That means the concentration of the diluted solution is:
0.136M / 4 =
<h3>0.034M HCl is the concentration of the diluted solution</h3>
Answer:
1. NaN₃(s) → Na(s) + 1.5 N₂(g)
2. 79.3g
Explanation:
<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>
NaN₃(s) → Na(s) + 1.5 N₂(g)
<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>
First, we have to calculate the moles of N₂ from the ideal gas equation.
The moles of NaN₃ are:
The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:
