Answer:
4) 2Na + Cl2 → 2NaCl
5) 2Mg + O2 → MgO
6) 2H2 + 2O2 → 2H2O
Explanation:
4) In order for it to be well balanced, yeah you've got to add 2 in front of Na before the reaction and after the reaction.
5) Add 2 in front of Mg bcuz of the same reason in number 4 but you don't have to add 2 in front of the others bcuz it's alrd well balanced.
6) Tbh for this one, it got me staring at my cat until he freaks out but like yeahh, I tried :')
Answer: In asexual reproduction, the genetic material of the parent cell doesn't combine with that from another cell. The simplest method of reproduction that algae employ is asexual binary fission, in which a cell splits into two, either at the equator or along its length.
Answer:
Therefore, the boiling points of the alkanes increase with molecular size. For isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors.
The boiling points of the normal alkanes increase with increasing molecular weight (Table 3.3). As the molecular weight increases, London forces increase because more atoms are present to increase the surface area or the molecules.
HOPE IT HELPS
PLEASE GIVE ME BRAINLIEST THANK YOU!♥︎
Answer:
a) Keq = 4.5x10^-6
b) [oxaloacetate] = 9x10^-9 M
c) 23 oxaloacetate molecules
Explanation:
a) In the standard state we have to:
ΔGo = -R*T*ln(Keq) (eq.1)
ΔGo = 30.5 kJ/moles = 30500 J/moles
R = 8.314 J*K^-1*moles^-1
Clearing Keq:
Keq = e^(ΔGo/-R*T) = e^(30500/(-8.314*298)) = 4.5x10^-6
b) Keq = ([oxaloacetate]*[NADH])/([L-malate]*[NAD+])
4.5x10^-6 = ([oxaloacetate]/(0.20*10)
Clearing [oxaloacetate]:
[oxaloacetate] = 9x10^-9 M
c) the radius of the mitochondria is equal to:
r = 10^-5 dm
The volume of the mitochondria is:
V = (4/3)*pi*r^3 = (4/3)*pi*(10^-15)^3 = 4.18x10^-42 L
1 L of mitochondria contains 9x10^-9 M of oxaloacetate
Thus, 4.18x10^-42 L of mitochondria contains:
molecules of oxaloacetate = 4.18x10^-42 * 9x10^-9 * 6.023x10^23 = 2.27x10^-26 = 23 oxaloacetate molecules