Answer:
polar orbit is one in which a satellite passes above or nearly above both poles of the body being orbited (usually a planet such as the Earth, but possibly another body such as the Moon or Sun) on each revolution. It has an inclination of about 60 - 90 degrees to the body's equator.[1] A satellite in a polar orbit will pass over the equator at a different longitude on each of its orbits.
Launching satellites into polar orbit requires a larger launch vehicle to launch a given payload to a given altitude than for a near-equatorial orbit at the same altitude, due to the fact that much less of the Earth's rotational velocity can be taken advantage of to achieve orbit. Depending on the location of the launch site and the inclination of the polar orbit, the launch vehicle may lose up to 460 m/s of Delta-v, approximately 5% of the Delta-v required to attain Low Earth orbit. Polar orbits are a subtype of Low Earth orbits with altitudes between 200 and 1,000 kilometers.[1]
Explanation:
Explanation:
It is known that 
value of acetic acid is 4.74. And, relation between pH and is as follows.
pH = pK_{a} + log ![\frac{[CH_{3}COOH]}{[CH_{3}COONa]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOH%5D%7D%7B%5BCH_%7B3%7DCOONa%5D%7D)
= 4.74 + log 
So, number of moles of NaOH = Volume × Molarity
= 71.0 ml × 0.760 M
= 0.05396 mol
Also, moles of 
= moles of 
= Molarity × Volume
= 1.00 M × 1.00 L
= 1.00 mol
Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.
Initial : 1.00 mol 1.00 mol
NaoH addition: 0.05396 mol
Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)
= 0.94604 mol = 1.05396 mol
As, pH = pK_{a} + log ![\frac{[CH_{3}COONa]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOONa%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 4.74 + log 
= 4.69
Therefore, change in pH will be calculated as follows.
pH = 4.74 - 4.69
= 0.05
Thus, we can conclude that change in pH of the given solution is 0.05.
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Volume= mass divided by density
V= m/d
55/3.23
= 17.03
Answer:
Explanation:
a. CuO+ 2HCl⇒CuCl2+ H2O
b. 
= 
= 0,05 (mol)
⇒
= =0,05 mol
⇒
= 0,05×135=6,75 (g)
c. 
=2× =0,1 (mol)
⇒
= 0,1×36,5= 3,65 (g)
⇒
= 
×100=36,5 (g)
⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=
×100=
×100≈ 16,67%
