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3 years ago

Rank the nonmetals in each set from most reactive (1) to least reactive (3). bromine chlorine iodine

Chemistry

2 answers:

Mice21 [21]3 years ago

7 0

Answer:

1. Chlorine 2. Bromine 3. Iodine

Explanation:

Anettt [7]3 years ago

3 0

Answer:

The answer is b

Explanation:

have common properties and are good conductors of heat and electricity. They reflect light and are malleable, and ductile.

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an unknown sample required 21.05 mL of 0.02047 M KmnO4 to reach the end point. How many moles of KmnO4 reacted?

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3 years ago

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Sveta_85 [38]

Answer:

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2 years ago

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