Answer:
What procedure did you use to complete the lab?
→Procedures that needs to be considered to complete the lab are- a thorough knowledge of lab assignments, knowledge about safety equipment, reviewing the MSDS of chemicals for lab experiment etc.
<h3>Explanation:</h3>
To be lab prepared one must follow these procedures-
1. One should have the knowledge of lab assignments to make the lab experiment easier.
2. To be aware about safety equipment and their uses in lab, like- the location of fire extinguisher in lab.
3. To know the steps of experiments to be performed
4. To fill notebook of lab with information regarding the experiment
5. One should review the data sheets of chemicals material safety.
6. To put on all the necessary dressings to perform experiment.
7. To have complete understanding about the experiment disposals.
Answer:
(a) The system does work on the surroundings.
(b) The surroundings do work on the system.
(c) The system does work on the surroundings.
(d) No work is done.
Explanation:
The work (W) done in a chemical reaction can be calculated using the following expression:
W = -R.T.Δn(g)
where,
R is the ideal gas constant
T is the absolute temperature
Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants
R and T are always positive.
- If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.
- If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.
- If Δn(g) = 0, W = 0, which means that no work is done.
<em>(a) Hg(l) ⇒ Hg(g)</em>
Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.
<em>(b) 3 O₂(g) ⇒ 2 O₃(g)
</em>
Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.
<em>(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g)
</em>
Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.
<em>(d) H₂(g) + F₂(g) ⇒ 2 HF(g)</em>
Δn(g) = 2 - 2 = 0. W = 0. No work is done.
Answer:
226.8 mg of mupirocin powder are required
Explanation:
Given that;
weight of standard pack = 22 g
mupirocin by weight = 2%
so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g
so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
mg of mupirocin powder are required = ?, lets rep this with x
Total weight of ointment = 22 + x g
Amount of mupirocin = 0.44 + x g
percentage of mupirocin in ointment is 3?
so
3/100 = 0.44 + x g / 22 + x g
3( 22 + x g ) = 100( 0.44 + x g )
66 + 3x g = 44 + 100x g
66 - 44 = 100x g - 3x g
97 x g = 22
x g = 22 / 97
x g = 0.2268 g
we know that; 1 gram = 1000 Milligram
so 0.2268 g = x mg
x mg = 0.2268 × 1000
x mg = 226.8 mg
Therefore, 226.8 mg of mupirocin powder are required
Answer:
The correct option is: B. 33%
Explanation:
Orbital hybridisation refers to the mixing of atomic orbitals of the atoms in order to form new hybrid orbitals. The concept of orbital hybridization is used to explain the structure of a molecule.
The sp² hybrid orbitals are formed by the hybridization of one 2s orbital and two 2p orbitals. <u>The three sp² hybrid orbitals formed have 33% s character and 67% p character.</u>
