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dangina [55]
2 years ago
14

What else should I add for creativity? Or else I won’t receive credit

Chemistry
2 answers:
padilas [110]2 years ago
4 0

Answer:

color the star with some spacey color(purpel, lavender, white spots)

You can also cover the rest of paper with different objects from space :)

Explanation:

Oxana [17]2 years ago
4 0

Answer:

More colors and add some planets.

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The solution that conducts electricity and has a pH value of 7 would most likely be a neutral solution. Water is among the best examples of a neutral solution. When the pH of a solution is considered to be lesser than 7, it is an acid, while if the pH is greater than 7, it is considered to be a basic solution.
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JOSE BOUGHT 750 BAGS OF PEANUTS FOR 375 HE INTENDS TO SELL EACH BAG FOR 0.15 MORE THAN HE PAID HOW MUCH SHOULD HE CHARGE FOR EAC
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Which type of precipitation can be considered acid rain?
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form of rain, snow, hail, dew, or fog that transports sulfur and. nitrogen compounds from the high atmosphere to the ground

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State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl
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Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

\Delta S = S(products)-S(reactants)----(1)

a) PCl3(l) + Cl2(g) \rightarrow PCl5(s)

Here  the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) 2HgO(s) \rightarrow 2Hg(l) + O2(g)

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) H2(g) \rightarrow 2H(g)

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) U(s) + 3F2(g) \rightarrow UF6(s)

Here  one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the  product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

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3 years ago
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