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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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Chemistry student needs of 55g acetone for an experiment. by consulting the crc handbook of chemistry and physics, the student d
sergeinik [125]

Answer:

             70.15 cm³

Solution:

Data Given;

                  Mass  =  55 g

                  Density  =  0.784 g.cm⁻³

Required:

                  Volume  =  ?

Formula Used:

                  Density  =  Mass ÷ Volume

Solving for Volume,

                  Volume  =  Mass ÷ Density

Putting values,

                  Volume  =  55 g ÷ 0.784 g.cm⁻³

                  Volume = 70.15 cm³

5 0
3 years ago
How can a flame test identify an unknown element?
Ilya [14]

Answer:

the answer for this question is the option D

6 0
3 years ago
A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes
ziro4ka [17]

Answer:

PCO2  = 0.6 25 atm

PSO2  = 1.2 75 atm

PO2 = 0.6  atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For  1 mole CS2 we need 3 moles O2  to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to  

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6  atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

8 0
3 years ago
Please help :) ill mark you as the brainiest hope you have a good day
Hitman42 [59]

Have a wonderful day :) thanks for the points

5 0
3 years ago
Please help me, Thank you!
babymother [125]

Answer:

amount of charge

Explanation:

Oxygen and sulfur are both in Group 16, which means they have a -2 charge. They have two more electrons than protons, making the charge of the ion negative.

Hope that helps.

4 0
3 years ago
Read 2 more answers
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