Part 1 :- Super gaint star have mass from 10 to 70 solar masses and brightness from 30,000 upto hundreds of thousand times the solar luminosity
blue super giant surface temperature is 20,000 to 50,000 degree Celsius example :0 Rigel its mass is around 20 times the sun mass it give light which 60,000 sun together give .
part 2 :- HR diagram : hertzsprung -Russel diagram ( attached below)
Answer:
ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).
Explanation:
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
