As the size of a star increases, luminosity increases. If you think about it, a larger star has more surface area. That increased surface area allows more light and energy to be given off. Temperature also affects a star's luminosity.
Answer:
2Na2O2+2H2O⟶O2+4NaOH
2×78g 32g
156g of Na2O2 produces 32g of O2,
12g of Na2O2 produces =15632×12=10.66g.
Density of O2 at NTP=1.428g/mL.
DensityMass=Vol.
1.42810.66=7.46mL
Vol. of O2 at NTP is 7.46mL.
Explanation:
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Answer:
pH = 10.505
Explanation:
Molar mass of Amphetamine ( C9H13N) = 135 g/mol
Given that the concentration of Amphetamine = 225 mg/L
mass of Amphetamine in one Liter = = 0.225 g
Number of moles of Amphetamine in one liter =
= 0.001667 mol
∴ molarity = 0.0017 M
C₉H₁₃N + H₂O --------> C₉H₁₃NH⁺ + OH⁻
I(M) 0.001667 M 0 0
C(M) -x x x
E(M) 0.001667 - x x x
Pkb = -log Kb = 4.2
∴ Kb = 6.309 x 10⁻⁵
Kb = 6.309 x 10⁻⁵
Equilibrium constant = [C₉H₁₃NH⁺][OH⁻]/ [C₉H₁₃N]
6.309 x 10⁻⁵ = x² / 0.001667-x
where 0.001667 -x ≅ 0.001667
Then;
x² = 6.309 x 10⁻⁵ × 0.001667
x² = 1.0517103 × 10⁻⁷
x =
x = 0.00032 M
x = [OH-] = 0.00032 M
∴ pOH = -log [OH-]
pOH = -log (0.00032)
pOH =3.495
pH = 14 - 3.495
= 10.505
Answer:
The difference in temperature is significant means that the lower-boiling liquid finishes distilling at a temperature that is too low for the higher-boiling liquid to be in vapor form yet.
Explanation:
The temperature will rise as the vapor of lower-boiling liquid rushes into the distillation head. However once the lower-boiling liquid is done distilling, there is a temperature drop because while the lower temperature liquid is done distilling, the temperature is still too low for the higher-boiling liquid to be rushing in as a vapour, so the temperature drops.
97.2g H20 x 1mole/18g of H2O x 6.022x10^23 molecules/1mole= 3.25 x10^24 molecules of H2O
hope this helps! :)