Answer:
Fat
Alkali
Explanation:
Fat and alkali are the two primary raw materials needed to manufacture soap.
Sodium hydroxide or potassium hydroxide is generally used as an alkali. The use of alkali depends on the intended application of the soap.
Raw animal fat was used in the past but these days, processed fat is used in the soap manufacturing process. Vegetable fats ( e.g, palm oil, olive oil, coconut oil) are also being used in soap manufacturing.
Additives are also used to enrich the color and texture of the soap.
Answer:
Solute concentration will afect the rate of a chemical reaction, because you must work with molarity
Explanation:
I think that solute mass may be it can affect the rate of reaction, if you have more mass in a solute, you will also have more moles.
If you want to know more, you have to consider temperature in the reaction and the presence of catalysts. They all, affect reactions.
Answer: D
Explanation:
A reducing agent is a species that reduces other compounds, and is thereby oxidized. The whole compound becomes the reducing agent. In other words, of a compound is oxidized, then they are the reducing agent. On the other hand, if the compound is reduced, it is an ozidizing agent.
Since we have established that a reducing agent is the compound being oxidized, we know that A is not our answer. An oxidized compound is losing electrons. Choice A states exactly this.
For B, this is true as we have established this already.
C is also correct. Since a reducing agent loses electrons, it becomes more positive. This makes the oxidation number increase.
D would be our correct answer. It is actually a good oxidizing agent is a metal in a high oxidation state, such as Mn⁷⁺.
When two waves come in contact with each other while traveling on the same medium, its called constructive interference. Constructive interference happens when two waves come together.
Pb(NO₃)₂ ⇒limiting reactant
moles PbI₂ = 1.36 x 10⁻³
% yield = 87.72%
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)
Required
- moles of PbI₂
- Limiting reactant
- % yield
Solution
Balanced equation :
Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
mol Pb(NO₃)₂ :
= 0.45 : 331 g/mol
= 1.36 x 10⁻³
mol NaI :
= 250 ml x 0.25 M
= 0.0625
Limiting reactant (mol : coefficient)
Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³
NaI : 0.0625 : 2 = 0.03125
Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)
moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)
Mass of PbI₂ :
= mol x MW
= 1.36 x 10⁻³ x 461,01 g/mol
= 0.627 g
% yield = 0.55/0.627 x 100% = 87.72%
