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Mariulka [41]
3 years ago
14

The boiling point of 100 milliliters of water at sea level, vapor pressure 760 mm hg is 100oc. if, under different conditions, t

he boiling point of the water is 98oc, what else changed?
a.the vapor pressure decreased.
b.the vapor pressure increased.
c.the volume of water decreased.
d.the time needed to boil the water decreased.
Chemistry
2 answers:
eimsori [14]3 years ago
7 0
It has to be D. the time needed to boil the water decreased

larisa [96]3 years ago
7 0

a) the vapor pressure decreased.

If 100 milliliters of water boiled at 98oC then the vapor pressure decreased. The boiling point of water changes if we vary the vapor (atmospheric) pressure. If the pressure goes down, the boiling point drops. The opposite is also true: higher pressure means a higher boiling point.

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A car at rest ends accelerates for 12 seconds. After this time the car is going 36 m/s. What was its acceleration?
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Acceleration is defined as velocity per unit time.

Acceleration=\frac{Velocity}{time}

a=\frac{dv}{dt}

Here, a=acceleartion,

v=velocity=36 m/s

t=time=12 s

a=\frac{dv}{dt}

a=\frac{36}{12}

a=3 ms⁻²

A car at rest ends accelerates for 12 seconds. After this time the car is going 36 m/s. So acceleration that is a=3 ms⁻².


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3 years ago
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¿Cuántos moles de C3H8 contienen 9.25 x 10^24 moléculas?
Igoryamba

Answer:

\large \boxed{\text{e. 15.4 mol C$_{3}$H}_{8}}

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A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found
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Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

\frac{1}{[A]} =\frac{1}{[A]_0} +kt

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so, the plot between 1/[A] and t is linear. So the statement b is true.

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d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

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