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Triss [41]
2 years ago
10

Lastly, Snape thinks we should try one more calculation. What is the retention factor if the distance travelled by the solvent f

ront is 2.00 cm, and the distance the ion is from the solvent front is 0.20 cm?
Hint: Be careful with how Snape worded this question.
Chemistry
1 answer:
bija089 [108]2 years ago
7 0

Answer:

The retention factor of an ion is 0.10 .

Explanation:

Retention factor is defined as ratio of distance of distance traveled by solute to the distance traveled by solvent on chromatogram.

R_f=\frac{d_{solute}}{d_{solvent}}

We have:

d_{solute}=d_{ion}=0.20 cm

d_{solven}=2.00 cm

The retention factor of an ion :

R_f=\frac{0.20 cm}{2.00 cm}=0.10

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The answer is (1) 10. The protons of Mg atom is 12. So the Mg atom has 12 electrons. The Mg2+ ion has lost two electrons so it has two positive charge. Then the answer is 10 electrons.
5 0
3 years ago
A gas has a volume of 6.35 L at 88.6 kPa. What will be the volume at standard pressure? Standard Pressure = 101.3 kPa
alukav5142 [94]

Answer:

5.55 L

Explanation:

This excersise can be solved by the Boyle's law.

This law for gases states that the pressure of a gas in a vessel is inversely proportional to the volume of the vessel.

P₁ . V₁  = P₂ . V₂

The law comes from the Ideal Gases Law, in the first term.

P . V = n . R . T  In this case, n . R . T are all constant.

6.35 L . 88.6 kPa = 101.3 kPa . V₂

V₂ = (6.35 L . 88.6 kPa) / 101.3 kPa

V₂ = 5.55 L

It is inversely proportional because, as it happened in this case, pressure was increased, therefore volume decreased.

5 0
3 years ago
Land forms on earth change everyday. Some are worn away and even destroyed. Some are built up and even formed anew. Processes th
kherson [118]
Missing part of the question. Please add which are the process to be classified.
5 0
2 years ago
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Identify the property of the matter described below.
ZanzabumX [31]

Answer: C.

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7 0
2 years ago
Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.
padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

brainly.com/question/13996238

#SPJ4

5 0
1 year ago
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