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3 years ago

A solution that has a relatively large amount of dissolved solute is called a(n) _______________________ solution.

Chemistry

2 answers:

OleMash [197]3 years ago

7 0

Ionic is the correct answer please let us know what you think

hoa [83]3 years ago

5 0

Answer:

Concentrated

Explanation: Imagine dissolving two cubes of sugar in a beaker containing 100ml of water, and dissolving five cubes of sugar in another beaker containing same volume of water. The beaker in which five cubes of sugar was dissolved in contains a more concentrated solution than the other.

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When 2 crystalline solid barium hydroxide octahydrate and crystalline solid ammonium nitrate are mixed in a beaker at room tempe

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Answer:

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3 years ago

Explain TWO ways that your body changes waste product levels to maintain an internal balance?

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3 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