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garri49 [273]
3 years ago
7

A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M,

what was the initial concentration?
Physics
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

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Half maximum velocity occurs at the point of half maximum kinetic energy which is exactly halfway down.

<h3>Conservation of energy</h3>

The principle of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.

M.A = K.E + P.E

At the maximum height, the kinetic energy of the ball while potential energy is maximum.

At the half-way down, the ball kinetic energy of the ball is equal to the potential energy.

Thus, half maximum velocity occurs at the point of half maximum kinetic energy which is exactly halfway down.

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2 years ago
If a 42kg rolling object slows from 11.5 m/s to 3.33 m/s how much work does friction do?
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Explanation:

work done by friction = 1/2 x 42 x ( 3.33^2 - 11.5^2)

= 21 ( 11.08 - 132.25)

= 21 ( - 121.17 )

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3 years ago
Jane puts some water into an electric kettle and then she connects it to the power source. She observes that after some time the
Setler [38]
C) electrical energy is transformed into heat energy
4 0
3 years ago
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saw5 [17]

The potential energy of the block is A) 490 J

Explanation:

The potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

It is calculated as follows:

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m is the mass of the object

g is the acceleration due to gravity

h is the height of the object above the ground

For the block in this problem, we have:

m = 10 kg

g=9.8 m/s^2

h = 5 m

Therefore, its potential energy is:

PE=(10)(9.8)(5)=490 J

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3 0
3 years ago
Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application
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F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

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