Answer:
Lithium diisopropylamide [(CH3)2CH]2NLi.
Explanation:
Lithium diisopropylamide (commonly abbreviated LDA) is a chemical compound with the molecular formula [(CH3)2CH]2NLi. It is used as a strong base and has been widely accepted due to its good solubility in non-polar organic solvents and non-nucleophilic nature. It is a colorless solid, but is usually generated and observed only in solution. It was first prepared by Hamell and Levine in 1950 along with several other hindered lithium diorganylamides to effect the deprotonation of esters at the α position without attack of the carbonyl group.
Answer:
c. 29 J
Explanation:
Step 1: Given data
- Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)
- Initial temperature: 22 °C
Step 2: Calculate the temperature change
ΔT = 37 °C - 22 °C = 15 °C
Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece
We will use the following expression.
Q = c × m × ΔT
Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J
The symbol of the radioactive nuclide, given the data is ⁴⁰₁₉K
<h3>Data obtained from the question</h3>
- Atomic number = 19
- Mass number = 40
- Symbol of nuclide =?
<h3>How to determine the nuclide</h3>
From the question given above, the atomic number of the nuclide is 19.
Comparing the atomic number (i.e 19) of the nuclide with those in the periodic table, the nuclide is potassium with a symbol of K
<h3>How to determine the symbol of the nuclide</h3>
- Atomic number (Z) = 19
- Mass number (A) = 40
- Name of nuclide = Potassium (K)
- Symbol of nuclide =?
The symbol of a nuclide is given as ᴬ₂X
Where
- A is the mass number
- Z is the atomic number
- X is the symbol of the element
Thus,
ᴬ₂X => ⁴⁰₁₉K
Therefore, the symbol of the nuclide is ⁴⁰₁₉K
Learn more about composition of atoms:
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Answer:
It is always less than the theoretical yield
Explanation:
For many chemical reactions, the actual yield is usually less than the theoretical yield. This is due to possible loss in the process or inefficiency of the chemical reaction.
Answer:
A. Adding another bulb will make the two of them light a little less brighter since the electrical flow is divided into two making them work.
B. There is a possibility that the flow of electricity maybe little due to the thickness of the wire. There could also be a rise in the temperature of the cell and the bulb.
C. The Bulb will shine brighter because of the extra power or electricity provided! Also, if the voltage is too much, expect the bulb to burn out or sadly, explode.
BTW GOODLUCK!!
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