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3 years ago

4. In a solution of sugar water, sugar is the ______. (SPS6a; DOK2)

Chemistry

2 answers:

Rzqust [24]3 years ago

6 0

Answer:

B. solute

Explanation:

lana [24]3 years ago

4 0

4. In a solution of sugar water, sugar is the ______. (SPS6a; DOK2)

a. solvent

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Which of the following would result if the gases that are responsible for the greenhouse effect were removed from our atmosphere

hichkok12 [17]

Answer:

Nighttime temperatures would be greatly reduced.

Explanation:

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3 years ago

The forensic technician at a crime scene has just prepared a luminol stock solution by adding 18.0 g of luminol to H2O creating

Bond [772]

Answer:

Part A. 1.355 mol/L

Part B. 0.100 mol

Part C. 74.0 mL

Explanation:

Part A.

The molar mass of luminol is 177.16 g/mol, so the number of moles at 18.0 g is:

n = mass/molar mass

n = 18.0/177.16

n = 0.1016 mol

The molarity is the number of moles divided by the volume (0.075 L)

C = 0.1016/0.075

C = 1.355 mol/L

Part B.

The number of moles is the molarity multiplied by the volume, so:

n = 5.00x10⁻² mol/L * 2.00 L

n = 0.100 mol

Part C.

To prepare a solution by dilution, we can use the equation

C1V1 = C2V2

Where C1 is the concentration of the initial (stock) solution, V1 is its volume necessary, C2 is the concentration of the diluted solution, and V2 is its volume.

Thus, C1 = 1.355 M, C2 = 0.05 M, V2 = 2.00 L

1.355V1 = 0.05*2

V1 = 0.074 L

V1 = 74.0 mL

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3 years ago

Do you think it is important to classify things

Dafna1 [17]

nononono JUST NO NOOOOOOOOO

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3 years ago

If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to ef

andrezito [222]

Answer:

Explanation:

Recall the law of effusion:

$\displaystyle \frac{r_1}{r_2} = \sqrt{ \frac{\mathcal{M}_2}{\mathcal{M}_1} }$

Because 5 mol of oxygen was effused in 10 seconds, the rate is 0.5 mol/s.

Let the rate of oxygen be r₁ and the rate of hydrogen be r₂.

The molecular weight of oxygen gas is 32.00 g/mol and the molecular weight of hydrogen gas is 2.02 g/mol.

Substitute and solve for r₂:

$\displaystyle \begin{aligned} \frac{(0.5\text{ mol/s})}{r_2} & = \sqrt{\frac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}} \\ \\ r_2 & = \frac{0.5\text{ mol/s}}{\sqrt{\dfrac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}}} \\ \\ & = 2.0\text{ mol/s}\end{aligned}$

Because there are 5 moles of hydrogen gas:

$\displaystyle 5.0\text{ mol} \cdot \frac{1\text{ s}}{2.0\text{ mol}} = 2.5\text{ s}$

In conclusion, it will take about 2.5 seconds for the hydrogen gas to effuse.

Check: Because hydrogen gas is lighter than oxygen gas, we expect that hydrogen gas will effuse quicker than oxygen gas.

6 0

2 years ago

Which of the following solutes has the greatest effect on the colligative properties for a given mass of pure water? Explain.

Strike441 [17]

Answer:

Option a.

0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i

Explanation:

To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)

These are the colligative properties:

ΔP = P° . Xm . i → Lowering vapor pressure

ΔT = Kb . m . i → Boiling point elevation

ΔT = Kf . m . i → Freezing point depression

π = M . R . T → Osmotic pressure

Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.

CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:

CaCl₂ → Ca²⁺ + 2Cl⁻

We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3

KNO₃ → K⁺ + NO₃⁻

We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2

Option a, is the best.

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3 years ago