The molarity of a salt solution made by dissolving 250.0 grams of NaCl in 775 mL of solution is <u>5.52 M</u> .
<u>Explanation:</u>
The molarity of a solution tells one how many moles of <u>solute</u> one can get <em>per liter</em> of solution.One should remember to must convert this volume to liters by using the conversion factor:
1 L = 10³ml
Sodium chloride (NaCl) has a molar mass of 58.44 g per mol , which means that the sample will contain:
This means that the molarity of the solution will be:
After the addition of 6.47 mL of 0.303 M HCl. Solution :- lets first calculate the moles of the NaOCl and HCl Moles = molarity *volume in liter Moles of NaOCl = 0.100 mol per L * 0.053 L =0.0053 mol Moles of HCl =0.303 mol per L *0.00647L=0.00196
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
Answer:
ΔG° = -533.64 kJ
Explanation:
Let's consider the following reaction.
Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)
The standard Gibbs free energy (ΔG°) can be calculated using the following expression:
ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)
where,
ni are the moles of reactants and products
ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products
ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)
ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)
ΔG° = -533.64 kJ
The mass of KCl per volume of solution=.35X(35.5+39)=26.075g/L
