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qaws [65]
3 years ago
7

The temperature program for a separation starts at a temperature of 50 °C and ramps the temperature up to 270 °C at a rate of 10

°C/minute. Which statement is NOT true for this separation?
A) At 10 °C/minute, a total of 22 minutes is needed to reach 270 oC.
B) Strongly retained solutes will remain at the head of the column while the temperature is low.
C) Weakly retained solutes will separate and elute early in the separation.
D) The vapor pressure of strongly retained solutes will increase as temperature increases.
E) Strongly retained analytes will give broad peaks.
Chemistry
1 answer:
Troyanec [42]3 years ago
6 0

Answer:

The correct answer to the question is Option E (Strongly retained analytes will give broad peaks).

Explanation:

The other options are true because:

A. Initial temp = 50 °C

   Final temp =  270 °C

Differences in temp = 270 - 50 = 220°C

Rate =  10 °C/minute.

So, at 10 °C/minute,

total of 220°C /10 °C = number of minutes required to reach the final temp.

220/10 = 22 minutes

B. A column has a minimum and maximum use temperature. Solutes that are already retained would remain stationary while temperatures are low. This would only change if there is an increase in temperature. Heat transfers more energy to the liquid which would make the solute interact with the column phase.

C. Weakly retained solutes may contain larger molecules, will separate by absorbing into the solvent early in separation making the mobile phase separates out into its components on the stationary phase.

D. Retained solute's vapor pressure is higher at higher temperatures making it possible for particle to escape more from the solute when the temperature is high than when it is low.

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The picture shows two containers filled with a gas.
aniked [119]

Answer: The average kinetic energy of the gas particles is greater in container B because it has a higher temperature.

3 0
2 years ago
Need asap thank you
cupoosta [38]

Answer:

c.boron-11

Explanation:

The atomic mass of boron is 10.81 u.

And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.

To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:

(10u)(x)+(11u)(1−x)100%=10.81u

Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.

Solving for x we get:

11u−ux=10.81u

0.19u=ux

x=0.19

1−x=0.81

And thus the abundance of boron-11 is roughly 81%.

6 0
2 years ago
Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. K2SO4 NH4I CoCl3
katrin [286]
The substance that releases the greatest amount of ions will have the greatest attractive forces within its solution, resulting in a reduced freezing point.
K₂SO₄ yields 3 ions
NH₄I yields 2 ions
CoCl₃ yields 4 ions

Freezing points:
CoCl₃ < K₂SO₄ < NH₄I
8 0
3 years ago
Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl
yuradex [85]

Answer:

You need 8,324 g of CaCl₂ yo make this solution

Explanation:

Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.

To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:

0,1500 L × \frac{0,500 mol}{L} = 0,075 CaCl₂ moles

Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:

0,075 CaCl₂ moles  × \frac{110,98 g}{mol} = 8,324 g of CaCl₂

So, you need <em>8,324 g of CaCl₂</em> to make 150,0 mL of a 0,500M solution

I hope it helps!

3 0
3 years ago
A system gains 687 kJ of heat, resulting in a change in internal energy of the system equal to 156 kJ. How much work is done?
Maslowich

Answer:

w = -531 kJ

1. Work was done by the system.

Explanation:

Step 1: Given data

  • Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
  • Change in the internal energy of the system (ΔU°): 156 kJ

Step 2: Calculate the work done (w)

We will use the following expression.

ΔU° = q + w

w = ΔU° - q

w = 156 kJ - 687 kJ

w = -531 kJ

By convention, when w < 0, work is done by the system on the surroundings.

4 0
3 years ago
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