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noname [10]
2 years ago
13

PLEASE HELP i’ll mark brainliest ( reporting fake answers)

Chemistry
1 answer:
kherson [118]2 years ago
6 0

Delta S reaction= Delta S products- Delta S reactants

don't forget to mulitiply by coefficients

also

here is a really slow way to do it

you know the moles of gas increased

so Delta S is positive

so its B or D

then just do the units digit to see which one match up

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noname [10]
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3 years ago
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The solubility of NaCH3CO2 in water is ~1.23 g/mL. What would be the best method for preparing a supersaturated NaCH3CO2 solutio
Len [333]

Answer:

b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.

Explanation:

The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).

<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>

<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.

<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.

<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.

5 0
2 years ago
What is the mole fraction, x, of solute and the molality, m (or b), for an aqueous solution that is 19.0% naoh by mass?
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We will assume that the solvent is water. So, if we have 100 grams of the solution, 19 grams will be sodium hydroxide, while the remaining 81 grams will be water.

The molar weight of sodium hydroxide, NaOH, is 40. The molar weight of water is 18. Finding the moles of each:

NaOH:
19 / 40 = 0.475

Water:
81 / 18 = 4.5

Total moles present:
4.5 + 0.475 = 4.975 moles

The mole fraction of NaOH is:
0.475 / 4.975 = 0.0955

The mole fraction of NaOH is 0.0955
5 0
2 years ago
The incomplete table below shows selected characteristics of gas laws.
lorasvet [3.4K]

Answer:

pressure and temperature

Explanation:

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Natali5045456 [20]

Answer:

The correct option is;

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