but heres a way to solve it
An athlete takes a deep breath, inhaling 1.85 L of air at 21°C and 754 mm Hg.
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How many moles of air are in the breath? How many molecules?
Gas constant, R= 8.314 J mol ¹ K-1
PV = nRT
PV
RT
h=
=
P
= 0.08206 L atm mol-1 K-1
= 62.36 L Torr mol-1 K-1 -
1 atm = 760 mm Hg = 760 Torr
754 Forr 1.85€
6236 Jerr 294K
Answer:
I don't know
Explanation:
I'm sorry I didn't learn this in my life,I have never reach in this level
Answer:
Sucrose (C12H22011) will not let a lightbulb light in water
Explanation:
Sucrose is too soluble to allow a lightbulb to light.
Answer:
Rate = k [X]⁻¹ [Z]²
Explanation:
[X] [Y] [Z] initial rate M M M M · s −1
Exp 1 0.30 0.20 0.35 0.210
Exp 2 0.60 0.10 0.70 0.420
Exp 3 0.60 0.20 0.70 0.420
Exp 4 0.60 0.40 0.35 0.105
In Experiment 2 and 3 where the concentrations of Y and Z were constant, doubling the concentration of Y had no effect on the rate of the reaction. This means, that the rate of the reaction is zero order with respect to Y.
In experiment 3 and 4, dividing the concentration of Z by 2, causes the rate of the reaction to decrease by 4. This means the rate of the reaction is second order with respect to Z.
In experiment 1 and 4, doubling the concentration of X, causes the rate of the reaction to decrease by half. This means that X has an order of -1 with respect to the rate of the reaction.
The rate expression is given as;
Rate = k [X]⁻¹[Y]⁰[Z]²
Rate = k [X]⁻¹ [Z]²
3. The difference between the potential energy of the products and the potential energy of reactants.
