Answer:
22:
Formular:

substitute:

23:
<em>Same</em><em> </em><em>element</em><em> </em><em>is</em><em> </em><em>represented</em><em> </em><em>by</em><em> </em><em>same</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>protons</em><em>.</em><em> </em>
Answer:
6 protons. 6 protons
7 neutrons. 8 neutrons
6 electrons. 6 electrons
Note: <u>Atoms</u><u> </u><u>with</u><u> </u><u>same</u><u> </u><u>proton</u><u> </u><u>number</u><u> </u><u>but</u><u> </u><u>different</u><u> </u><u>mass</u><u> </u><u>number</u><u> </u><u>are</u><u> </u><u>called</u><u> </u><u>isotopes</u>
A.) Delta . Delta is a landform that forms from Deposition.
Answer:
The reaction that is used to join fatty acid and glycerol is known as Lipogenesis.
Explanation:
Lipogenesis is an important anabolic pathway that helps in the biosynthesis of triacylglycerol by joining glycerol with fatty acid by ester linkage.Lipogenesis occur in liver and adipose tissue .
Lipogenesis takes place in our body to store excess fatty acid in form of triacylglycerol which is a complex lipid molecule.
The first three quantum numbers for electrons located in subshell 4s are :
Answer:
10043.225 J
Explanation:
We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:
Mass (m) = 15.5g
Latent heat of fussion of water (L) = 334J/g
Heat (Q1) =..?
Q1 = mL
Q1 = 15.5 x 334
Q1 = 5177 J
Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.
This is illustrated below:
Mass = 15.5g
Initial temperature (T1) = 0°C
Final temperature (T2) = 75°C
Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C
Specific heat capacity (C) of water = 4.186J/g°C
Heat (Q2) =?
Q2 = MCΔT
Q2 = 15.5 x 4.186 x 75
Q2 = 4866.225 J
The overall heat energy needed is given by:
QT = Q1 + Q2
QT = 5177 + 4866.225
QT = 10043.225 J
Therefore, the amount of energy required is 10043.225 J