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3 years ago

Which pair of elements is MOST likely to chemically combine and form ionic bonds?

1 answer:

dsp733 years ago

5 0

The right answer for the question that is being asked and shown above is that: "C. carbon and hydrogen." The pair of elements is MOST likely to chemically combine and form ionic bonds are carbon and hydrogen

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During early research in nuclear energy, why were scientists surprised that the products from splitting a uranium nucleus had le

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Scientists expected that the law of conservation of mass would apply to nuclear fission in terms of the masses of the subatomic particles. In reality, the mass of an atom is not equal to the sum of the masses of the subatomic particles that make it up. This is because of the energy that binds the subatomic particles. This energy has mass and when the bond is broken, the mass of the energy of the bonds is lost resulting to what we now cal, a mass defect.

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Simplify the numercial expression 13-2x5

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Explanation:

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2 years ago

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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

Strike441 [17]

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