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sveticcg [70]
3 years ago
14

List two detection (i.e. visualisation) techniques commonly used to visualise compounds in TLC.

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
7 0

Answer:

The most common non-destructive visualization method for TLC plates is ultraviolet (UV) light. A UV lamp can be used to shine either short-waved (254nm) or long-waved (365nm) ultraviolet light on a TLC plate with the touch of a button

Explanation:

hope this helps

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A researcher studying the nutritional value of a new candy places a 4.90 g sample of the candy inside a bomb calorimeter and com
snow_lady [41]

Answer:

449730.879 cal/g

Explanation:

Given data:

Mass of sample = 4.9 g

Change in temperature  = 2.08 °C  (275.23 k)

Heat capacity of calorimeter = 33.50 KJ . K⁻¹

Solution:

C(candy) = Q/m

Q = C (calorimeter) × ΔT

C(candy) = C (calorimeter) × ΔT / m

C(candy) =  33.50 KJ . K⁻¹ × 275.23 K / 4.90 g

C(candy) = 9220.205 KJ / 4.90 g

C(candy) =  1881.674 KJ / g

It is known that,

1 KJ /g = 239.006 cal/g

1881.674 × 239.006 = 449730.879 cal/g

8 0
3 years ago
Hydrogen has three isotopes, 1H, 2H, and 3H. What is the difference between these three isotopes?
Firdavs [7]
C... The number of neutrons ranges from 1H with 0 neutron, 2H with 1 neutron and 3H with 2 neutrons.
6 0
3 years ago
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In a periodic table, dots around an elements symbol indicate the number of valence electrons in an atom ? True Or False
Liono4ka [1.6K]
True,I think.(so don't really count on my answer sorry :D)
6 0
2 years ago
Describe the three values in a learner-centred curriculum that a teacher can use
ad-work [718]

Answer:

to talk in a way that the class can all understand . help them in thing they are falling . and have a good relashaship.

Explanation:

3 0
2 years ago
A reaction vessel is charged with hydrogen iodide, which partially decomposes to molecular hydrogen and iodine: 2HI (g) H2(g) +
leonid [27]

Answer:

The value of the equilibrium constant: K_{p} = 0.25

Explanation:

Given reaction: 2HI (g) ⇌ H₂(g) + I₂(g)

Number of moles of- HI: n₁ = 2 mole; H₂: n₂ = 1 mole; I₂: n₃ = 1 mole

Total number of moles: n = n₁ + n₂ + n₃ = 2 + 1 + 1 = 4 moles

The equilibrium constant for the given reaction is given as:

K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}

Given: Temperature: T = 425 °C = 425 + 273 = 698 K

The partial pressure: pHI = 0.708 atm,

and, pH₂ = pI₂

 

∵ <em>partial pressure of a given gas</em>: pₐ = Χₐ . P

Here, P is the total pressure

Χₐ is the <em>mole fraction</em> of given gas and is given by the equation

\chi_{a} = \frac{number \, of \,moles \,of \,given \,gas (n_{a})}{total \,number \,of \,moles (n)}

Mole fraction for HI: \chi_{1} = \frac {n_{1}}{n} = \frac {2}{4} = 0.5

Mole fraction for H₂: \chi_{2} = \frac {n_{2}}{n} = \frac {1}{4} = 0.25

Mole fraction for I₂: \chi_{3} = \frac {n_{3}}{n} = \frac {1}{4} = 0.25

Thus, Χ₂ = Χ₃ = 0.25

The partial pressure of HI is given by;

pHI = Χ₁ P

0.708 atm = 0.5 × P

⇒ P = 1.416 atm

 

As the partial pressures: pH₂ = pI₂

∴ pH₂ = pI₂ = Χ₂ P = Χ₃ P = 0.25 × 1.416 atm = 0.354 atm

Therefore, the value of Kp can be calculated as:

K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}

K_{p} = \frac{0.354 atm\times 0.354 atm}{(0.708 atm)^{2}} = 0.25

<u>Therefore, the value of the equilibrium constant: </u>K_{p} = 0.25

8 0
3 years ago
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