Answer:
a) 38.2 % mass
b) 61.8 g solute/100 g solvent
c) 1.65 g/mL
Explanation:
Given the data:
mass of solute = 17.5 g
mass of solvent= 28.3 g
total solution volume= 27.8 mL
a)- mass percent= mass of solute/mass of solution x 100
mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g
mass % = 17.5 g/45.8 g x 100 = 38.2 % mass
b)- solubility = grams of solute/ 100 g solvent
= 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent
c)- density = massof solution/total volumesolution = 45.8 g/27.8 mL = 1.65 g/mL
Answer:
<u>1. Net ionic equation:</u>
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s)
<u />
<u>2. Volume of 1.0M AgNO₃</u>
Explanation:
1. Net ionic equation for the reaction of NaCl with AgNO₃.
i) Molecular equation:
It is important to show the phases:
- (aq) for ions in aqueous solution
- (s) for solid compounds or elements
- (g) for gaseous compounds or elements
- NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)
ii) Dissociation reactions:
Determine the ions formed:
- NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
- AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
- NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)
iii) Total ionic equation:
Substitute the aqueous compounds with the ions determined above:
- Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)
iv) Net ionic equation
Remove the spectator ions:
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer
2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl
i) Determine the number of moles of AgNO₃
The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl
The number of moles of AgCl is determined using the molar mass:
- number of moles = mass in grams / molar mass
- molar mass of AgCl = 143.32g/mol
- number of moles = 5.84g / (143.32g/mol) = 0.040748 mol
ii) Determine the volume of AgNO₃
- molarity = number of moles of solute / volume of solution in liters
- V = 0.040748mol / (1.0M) = 0.040748 liter
- V = 0.040748liter × 1,000ml / liter = 40.748 ml
Round to two significant figures: 41ml ← answer
2 C2H2 + 5 02 > 4 CO2 + 2 H2O
Products - Reactants ( all units are kJ/mo1):
(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1
-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1
answer: -1255.6 kJ/mo1
As number of gaseous moles in reactant and prodict are same that is 4
So No change will occur
Answer:
The coefficient of O2 is 11
Explanation:
Step 1:
The equation for the reaction:
FeS2 + O2 → SO2 + Fe2O3
Step 2:
Balancing the equation. The equation can be balance as follow:
FeS2 + O2 → SO2 + Fe2O3
There are 2 atoms of Fe on the right side and 1 atom on the left. It can be balance by putting 2 in front of FeS2 as shown below:
2FeS2 + O2 → SO2 + Fe2O3
There are 4 atoms of S on the left side and 1 atom on the right side. It can be balance by putting 4 in front of SO2 as shown below:
2FeS2 + O2 → 4SO2 + Fe2O3
Now, there are a total of 11 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 11/2 in front of O2 as shown below:
2FeS2 + 11/2O2 → 4SO2 + Fe2O3
Multiply through by 2 to clear the fraction as shown below:
4FeS2 + 11O2 → 8SO2 + 2Fe2O3
Now the equation is balanced.
The coefficient of O2 is 11
