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E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
a) M(N</span>₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, <span>CO and N</span>₂ <span>have iexact molecular masses and will effuse at nearly identical rates.</span>
Answer: 12.78ml
Explanation:
Given that:
Volume of KOH Vb = ?
Concentration of KOH Cb = 0.149 m
Volume of HBr Va = 17.0 ml
Concentration of HBr Ca = 0.112 m
The equation is as follows
HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)
and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)
Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb
(0.112 x 17.0)/(0.149 x Vb) = 1/1
(1.904)/(0.149Vb) = 1/1
cross multiply
1.904 x 1 = 0.149Vb x 1
1.904 = 0.149Vb
divide both sides by 0.149
1.904/0.149 = 0.149Vb/0.149
12.78ml = Vb
Thus, 12.78 ml of potassium hydroxide solution is required.
avogadros law states that under constant temp n pressure, volume of a gas is directly proportional to amount of gas
amount of gas1 = 352millimoles
amount of gas2 = 352+100 = 452millimoles
new vol = old vol x 452/352 = 25.2 x 452/352
= 32.4 mL
