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2 years ago

15

Which process is spontaneous? electrolysis of water into oxygen and hydrogen spreading of a strong fragrance of perfume in a roo

m separation of sodium and chlorine from sodium chloride removal of iron fillings from a mixture by using a magnet?

1 answer:

dmitriy555 [2]2 years ago

3 0

Yes I think oxides I's the answer

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Homolysis, or homolytic bond dissociation, produces a very specific type of product under certain reaction conditions. In Part 1

vekshin1

Answer:

1) R₃CO , H, H₃C, a carbon free radical

2) high temperature, ultraviolet irradiation

Explanation:

1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :

R₃CO

H

H₃C

a carbon free radical

2) Homolysis require high temperature, ultraviolet irradiation.

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3 years ago

Figure 1 shows a seaside cliff. Figure 2 shows the same cliff after a period of time has passed. What caused the change from fig

NeX [460]

The change from figure one to figure two was most likely caused by erosion. Erosion is the process of something being eroded by wind, water, or other natural agents.

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2 years ago

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The isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 min at 500°c. the time it takes f

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<span>As mentioned, the isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 min at 500°c. A first-order reaction kinetic rates means that the rate is constant throughout the reaction.

Thus, the time it takes for the partial pressure of cyclopropane to decrease from 1 atm to 0.125 atm at 500°c is </span><span>57 minutes.</span>

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3 years ago

What is the main basis for the classification system they chose?

VARVARA [1.3K]

What classification system????

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3 years ago

When 131.0 mL of water at 26.0°C is mixed with 81.0 mL of water at 85.0°C, what is the final temperature? (Assume that no heat i

JulsSmile [24]

Answer:

63.52°C is the final temperature

Explanation:

1) 131.0 mL of water at 26.0°C

Mass of water = m

Volume of the water =131.0 mL

Density of the water = 1.00 g/mL

$Density=1.00 g/mL=\frac{m}{131.0 mL}$

m = 131.0 g

Initial temperature of the water = $T_i$ = 26.0°C

Final temperature of the water = $T_f$

Change in temperature , $\Delta T=T_f-T_i$

Heat absorbed 131.0 g of water = Q

$Q=m\times c\times \Delta T$

2) 81.0 mL of water at 85.0°C

Mass of water = m'

Volume of the water =81.0 mL

Density of the water = 1.00 g/mL

$Density=1.00 g/mL=\frac{m'}{81.0 mL}$

m' = 81.0 g

Initial temperature of the water = $T_i'$ = 85.0°C

Final temperature of the water = $T_f'$

Change in temperature , $\Delta T'=T_f'-T_i'$

Heat lost by 81.0 g of water = Q'

$Q'=m'\times c\times \Delta T'$

After mixing both liquids the final temperature will become equal fro both liquids.

$T_f=T_f'$

Since, heat lost by the water at higher temperature will be equal to heat absorbed by the water at lower temperature.

Q=-Q' (Law of conservation of energy.)

Let the specific heat of water be c

$m\times c\times \Delta T=m'\times c\times \Delta T'$

$131.0 g\times c(T_f-26^oC)=-(81.0.0 g\times c(T_f-85^oC))$

$T_f=63.52^oC$

63.52°C is the final temperature

4 0

3 years ago