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Aneli [31]
3 years ago
8

Based on conductivity values, classify the compounds in each set of solutions as molecular, ionic, or acids. what is the level o

f dissociation (complete, varying, or none) of each set
Chemistry
1 answer:
DochEvi [55]3 years ago
8 0
Electrolytic conductivity is measure of ability of solution to conduct electricity. It is also as specific conductance. Electrolytic conductivity is different as compared to electrical conductivity. Electrolytic conductivity increase with temperature, while electrical conductivity decreases with increasing temperature. Further, electrolytic conductivity occurs because of presence of ions in solutions, while electrical conductivity occurs due to presence of valence electrons.

In current context, number of ions present in solutions depends on dissociation constant values. For molecules, which donot under dissociation in solvent, no electrolytic conductivity will be observed. Generally, ionic compounds undergo complete dissociation in solution, hence they will display maximum conductance. In case of acids, conductivity values will vary depends upon dissociation constant values. Weak acid are characterized by lower dissociation constant values, hence they have lower conductivity as compared to strong acids.  <span />
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3 years ago
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 1.92 g of sodium
Oksana_A [137]

Answer:

The % yield is 27.0 %

Explanation:

<u>Step 1: </u>Data given

Mass of sulfuric acid = 4.9 grams

Mass of sodium hydroxide = 7.8 grams

Mass of sodium sulfate produced = 1.92 grams

Molar mass H2SO4 = 98.08 g/mol

Molar mass NaOH = 40 g/mol

Molar mass Na2SO4 = 142.04 g/mol

<u>Step 2: </u>The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

<u>Step 3</u>: Calculate moles H2SO4

Moles H2SO4 = Mass H2SO4 / Molar mass H2SO4

Moles H2SO4 = 4.9 grams / 98.08 g/mol =

Moles H2SO4 = 0.05 moles

<u>Step 4:</u> Calculate moles NaOH

Moles NaOH = 7.8 grams / 40 g/mol

Moles NaOH = 0.195 moles

<u>Step 5</u>: Calculate limiting reactant

For 1 mole H2SO4 consumed ,we need 2 moles NaOH to produce 1 mole Na2SO4 and 2 moles H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.05 moles).

NaOH is in excess. There will react 2*0.05 = 0.1 moles

There will remain 0.195 -0.1 = 0.095 moles NaOH

<u>Step 6:</u> Calculate moles Na2SO4

For 1 mole H2SO4 consumed ,we need 2 moles NaOH to produce 1 mole Na2SO4

For 0.05 moles H2SO4, we have 0.05 moles Na2SO4

<u>Step 7:</u> Calculate mass of Na2SO4

Mass Na2SO4 = Moles Na2SO4 * Molar mass Na2SO4

Mass = 0.05 moles * 142.04 g/mol = 7.102

This is the theoretical yield

<u>Step 8:</u> Calculate the percent yield of Na2SO4

% yield = (actual yield / theoretical yield) * 100%

% yield = (1.92 /  7.102) *100% = 27.0 %

The % yield is 27.0 %

7 0
3 years ago
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