Answer:
The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%
Explanation:
When a solute dissolves in a solvent, the mass of the resulting solution is a sum of the mass of the solute and the solvent.
A percentage is a way of expressing a quantity as a fraction of 100. In this case, the percentage by mass of a solution is the number of grams of solute per 100 grams of solution and can be represented mathematically as:
In this way it allows to precisely establish the concentration of solutions and express them in terms of percentages.
In this case:
- mass of solute: 3.55 g
- mass of solution: 3.55 g + 88 g= 91.55 g
Replacing:
Percent by mass= 3.88%
<u><em>The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%</em></u>
F₂ + 2 NaI → 2 NaF + I₂
<span>It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F</span>₂ is present<span>)</span>
Answer:
The answer to your question is: letter D
Explanation:
In a combustion reaction, the reactants are always a molecule with Carbon that reacts with oxygen and the products are carbon dioxide and water.
According to the explanation, the only possible solution is:
a) C₆H₁₂O₂(l) ⇒ 6 C(s) + 6 H₂(g) + O₂(g)
b) Mg(s) + C₆H₁₂O₂(l) ⇒ MgC₆H₁₂O₂(aq)
c) 6 C(s) + 6 H₂(g) + O₂(g) ⇒ C₆H₁₂O₂(l)
d) C₆H₁₂O₂(l) + 8 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(g)
e) None of the above represent the combustion of C₆H₁₂O₂.
Molar mass O2 = 31.99 g/mol
Molar mass CO2 = 44.01 g/mol
Moles ratio:
<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2
mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01
mass of O2 = 8174.8575 / 132.03
mass of O2 = 61.916 g
Therefore:
1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916
moles O2 = 61.916 x 1 / 31.99
moles = 61.916 / 31.99 => 1.935 moles of O2
