The charge on a magnesium ion is +2. When it formed an ionic compound with another element, it lost

Chemistry

1 answer:

strojnjashka [21]3 years ago

4 0

I believe it’s loses two electrons.

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What is the sign of △S for each of the following processes? I. The separation of gaseous molecules of UF6, into 238UF6 and 235UF

gavmur [86]

Answer:

Explanation:

Entropy is the degree of randomness or disorderliness of as system. The more random a system tends, the more positive the entropy. Gases have the highest entropy whereas solids have little to no entropy.

When a reaction goes from one initally with a lower entropy to one with a higher entropy, the change in entropy, ΔS is postive. ΔS is negative when a reaction goes from one with a higher entropy to one with lower entropy.

When there is no entropy change, both the reactants and products are in the same state.

In the first reaction, UF₆ is in a gaseous phase and it was seperated into two gaseous products, 238-UF₆ and 235-UF₆. Therefore, ΔS is zero i.e no change in entropy of the system.

In the second reaction, solid I₂ is dissolved in CCl₄. The solid substance goes into solution to form a liquid medium. This is an increase in the randomness or entropy of the system. Therefore ΔS is +ve.

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3 years ago

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dalvyx [7]

Answer:

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3 years ago

For parts a & b below, derive only the initial value problem set up.

Otrada [13]

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

$\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )$

The rate of the sugar coming into the tank is 0

$\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min$

$So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100} \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C \\ \\ \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}$