Answer:
Explanation:
In a simple cubic lattice lattice, the atoms are present at the eight corners of the cibe.
Since it is mentioned that the fluorine is present at the corners and also 1 corners are shared by 8 unit cells. So, share of atom in one unit cell is:- 
Also, the metal, M occupy half of the body centre. A cube has only one body centre. So, share of M in each unit cell:- 
Thus, the formula is:-
Or simplifying
Water acts as solvent and it has no effect on the amount of the acid or the titrating agent.
<span>Also, the concentration of the titrating agent must be known exactly, the amount of solvent is less important (the more diluted it is, the more accurate is the titration).</span>
Answer:
The valid quantum numbers are l=0, l=-2 and l= 2.
Explanation:
Given that,
n = 3 electron shell
Suppose, the valid quantum numbers are
l = 3
m = 3
l = 0
m = –2
l = –1
m = 2
We know that,
The value of n = 3
Principle quantum number :
Then the principal quantum number is 3. Which is shows the M shell.
So, n = 3
Azimuthal quantum number :
The azimuthal quantum number is l.
Magnetic quantum number :
The magnetic quantum number is
Hence, The valid quantum numbers are l=0, l=-2 and l= 2.
<h3><u>Answer;</u></h3>
1. In the light reactions, light energy is used to oxidize H2O to O2.
2. The electrons derived from this oxidation reaction in the light reactions are used to reduce NADP+ to NADPH.
3. The Calvin cycle oxidizes the light-reactions product NADPH to NADP+.
4. The electrons derived from this oxidation reaction in the Calvin cycle are used to reduce CO2 to G3P.
<h3><u>Explanation;</u></h3>
- <em><u>In the light reactions, light energy is used to remove electrons from (oxidize) water, producing O2 gas. These electrons are ultimately used to reduce NADP+ to NADPH.
</u></em>
- In the Calvin cycle, NADPH is oxidized back to NADP+ (which returns to the light reactions). The electrons released by the oxidation of NADPH are used to reduce three molecules of CO2 to sugar (G3P), which then exits the Calvin cycle.
- As ATP and NADPH are used in the Clavin cycle, they produce ADP and NADP+, respectively, which are returned to the light reactions so that more ATP and NADPH can be formed.
The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV
Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules
From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s
