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saw5 [17]
3 years ago
10

Determine whether each melting point observation corresponds to a pure sample of a single compound or to an impure sample with m

ultiple compounds.
a. Wide melting point range __________
b. Experimental melting point is close to literature value _________
c. Experimental melting point is below literature value _________
d. Narrow melting point range __________
Chemistry
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer:

Wide melting point range - impure sample with multiple compounds

Experimental melting point is close to literature value - pure sample of a single compound

Experimental melting point is below literature value - impure sample with multiple compounds

Narrow melting point range - pure sample of a single compound

Explanation:

The melting point of substances are easily obtainable from literature such as the CRC Handbook of Physics and Chemistry.

A single pure substance is always observed to melt within a narrow temperature range. This melting temperature is always very close to the melting point recorded in literature for the pure compound.

However, an impure sample with multiple compounds will melt over a wide temperature range. We also have to recall that impurities lower the melting point of a pure substance. Hence, the experimental melting point of an impure sample with multiple compounds is always below the literature value.

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If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 Liters, what is the temperature in Kelvin(K)
kvv77 [185]

Answer:

205 K (to 3 significant figures)

Explanation:

Assuming that 4 moles of the gas behaves like an ideal gas and obey the kinetic molecular theory.

Let's apply the ideal gas law, pV= nRT.

Here p denotes the pressure of the gas, V is for volume, n is the number of moles of the gas, R is the universal gas constant and T is the temperature.

Substitute the given information into the equation:

5.6 atm ×12 L= 4 mol ×R ×T

Since pressure is in atm and volume is in L, we can use R= 0.08206 L atm K⁻¹ mol⁻¹.

5.6 atm ×12 L= 4 mol ×0.08206 L atm K⁻¹ mol⁻¹ ×T

T= 67.2 ÷0.32824

T= 204.73 (5 s.f.)

T= 205 K (3 s.f.)

4 0
2 years ago
Methane undergoes combustion. Which products form?
Aleksandr [31]

<u>M</u><u>e</u><u>t</u><u>h</u><u>a</u><u>n</u><u>e</u><u> </u>is a carbon compound which undergoes combustion to <em><u>release energy</u></em> and form bi production which are <u>Carbon</u><u> </u><u>dioxide</u><u> </u>( CO2 )<u> </u><u>and</u><u> </u> <u>W</u><u>ater</u> ( H20 ).

the balanced chemical equation for the reaction is : -

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3 0
3 years ago
What is carboxylic acid?​
maksim [4K]

Answer:

Carboxylic acid

A carboxylic acid is an organic acid that contains a carboxyl group attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO₂H, with R referring to the alkyl, alkenyl, aryl, or other group. Carboxylic acids occur widely. Important examples include the amino acids and fatty acids.

7 0
2 years ago
Magnesium (average atomic mass = 24.3050 amu) consists of three isotopes with masses 23.9850 amu, 24.9858 amu, and 25.9826 amu.
iris [78.8K]

Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%

Explanation: Given that the average atomic mass(M) of magnesium

= 24.3050amu

Mass of first isotope (M1) = 23.9850amu

Mass of middle isotope (M2)=24.9858amu

Mass of last isotope(M3)= 25.9826amu

Total abundance = 1

Abundance of middle isotope = 0.10

Let abundance of first and last isotope be x and y respectively.

x+0.10+y =1

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M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope

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Substitute x= 0.90-y

Then

y = 0.11

Since y=0.11, then

x= 0.90-0.11

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Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%

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Why do plants need sap
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7 0
3 years ago
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