Answer:
0.6941 mg
Explanation:
First we <u>calculate how many LiNO₃ moles there are</u>, using the <em>given concentration and volume</em>:
- 1.0 mL * 0.10 M = 0.10 mmol LiNO₃
As 1 mol of LiNO₃ contains 1 mol of Li,<em> in the problem solution there are 0.10 mmol of Li</em> (the only metallic ion present).
Now we<u> convert Li milimoles into miligrams</u>, using its <em>atomic mass</em>:
- 0.10 mmol Li * 6.941 mg/mmol = 0.6941 mg
Answer:
First confirm the reaction is balanced:
C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).
a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.
b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).
There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.
MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g
c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.
The other questions use the same technique and will give you some much needed practice.
Explanation:
The graph is not given in the question, so, the required graph is attached below:
Answer:
According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.
The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.
