Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
Li is in group 1 so it has same number of valence electrons as hydrogen.
hope it helps.
The only one that I can do without google is 47. Sorry that I can't answer the others. The answer to 47 is this: you know that the western side of the hill has the steepest slope because the ovals showing altitude are way closer together. The closer the circles/ovals are, the steeper the slope is.
Sorry if this doesn't help much, but I answered what I could without cheating.
Foxeslair
Answer:
y1 = 0.3162
y2 = 0.6838
Explanation:
ok let us begin,
first we would be defining the parameters;
at 25°C;
1-propanol P1° = 20.90 Torr
2-propanol P2° = 45.2 Torr
From Raoults law:
P(1-propanol) = P⁰ × X(1-propanol)
P(1-propanol) = 20.9 torr × 0.45 = 9.405
P(1-propanol) = 9.405 torr
Also P(2-propanol) = P⁰ × X(2-propanol)
P(2-propanol) = 45.2 torr × 0.45
P(2-propanol) = 20.34 torr
but the total pressure = sum of individual pressures
total pressure = 9.405 + 20.34
total pressure = 29.745 torr
given that y1 and y2 represent the mole fraction of each in the vapor phase
y1 = P1 / total pressure
y1 = 9.405/29.745
y1 = 0.3162
Since y1 + y2 = 1
y2 = 1 - y1
∴ y2 = 1 - 0.3162
y2 = 0.6838
cheers, i hope this helps.
I'm pretty sure its D, Time of discovery of the body. Let me know (:
