Oran fills in the table below to organize information about the gas laws.

What would the final freezing point of water be if 3 mol of sugar were added to 1 kg of water (Kf = 1.86C/(mol/kg) for water and

Fudgin [204]

is this for a test or are you genuinely interested? molality = mols sugar/kg solvent

Solve for molality

delta T = Kf*m

Solve for delta T and subtract from zero C to find the new freezing point.

or

-5.58

4 0

2 years ago

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6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ

qwelly [4]

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy ( $\Delta G$ ) should be negative.

We know, $\Delta G=\Delta H-T\Delta S$ , where T is temperature in Kelvin scale.

Reaction A: $\Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{10.5\times 10^{3}}{30}$

or, $T> 350$

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: $\Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{1.8\times 10^{3}}{-113}$

or, $T> -16$

So at a temperature greater than -16 K, the reaction is spontaneous.

3 0

2 years ago

When salt is introduced to water, the temperature at which freezing occurs is?

Vladimir [108]

28.4 degrees fahrenheit

5 0

2 years ago

Which describes what a velocity/time graph would look like with no acceleration?

Gwar [14]

Answer:

Explanation:With no acceleration the graph would be flat.

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2 years ago

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A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

2 years ago