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ella [17]
3 years ago
15

What is the molarity of a solution prepared by dissolving 12.0 g of sodium acetate, nac2h3o2, in water and diluting it to 250.0

ml?
Chemistry
1 answer:
amid [387]3 years ago
3 0
Molarity is defined as the number of moles of solute in 1 L of solution 
the mass of sodium acetate added - 12.0 g 
molar mass of sodium acetate - 82 g/mol
number of moles = mass present / molar mass
number of moles of sodium acetate added = 12.0 g / 82 g/mol = 0.146 mol
the number of moles in 250.0 mL - 0.146 mol
therefore number of moles in 1000 mL - 0.146 mol / 250.0 mL x 1000 mL/L = 0.584 mol
there are 0.584 mol in 1 L - 0.584 mol/L
the molarity of sodium acetate is 0.584 M
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mario62 [17]

Answer:

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Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

<u>a. C₂H₄:</u>

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Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

<u>b. C₂H₂:</u>

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The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

<u>с. С₃Н₈</u>

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The mole ratio is 1 mol C₃H₈ : 5 mol O₂

<u>d. C₂H₆</u>

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7 0
4 years ago
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Answer:

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Explanation:

¡Hola!

En este caso, de acuerdo con el concepto de sal, la cual está generalmente dada por la presencia de al menos un metal y un no metal, es posible encontrar cuatro tipos de estas; hidrácidas, oxácidas, básicas y ácidas, en las que las primeras dos son neutras pero la segunda tiene presencia de oxígeno, la tercera tiene iones hidróxido adicionales y la cuarta iones hidrógeno de más.

Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:

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¡Saludos!

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