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a_sh-v [17]
3 years ago
6

!NO LINKS!

Chemistry
1 answer:
ad-work [718]3 years ago
8 0

A sodium atom is more likely to react with a chlorine atom than a potassium atom.

The atoms of elements in the periodic table are arranged in groups and periods. The elements in the same group of the periodic table have similar reactivity.

We know that metals do not often combine with each other except in alloys. Metals of group 1 such as sodium and potassium do not form alloys hence they do not combine.

On the other hand, the electronegativity difference between sodium and chlorine is high. This causes the both of them to easily combine and form and ionic compound.

Learn more: brainly.com/question/11527546

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When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
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When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

<u>No of moles = mass of the substance÷molar mass of the substance</u>

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

<u>Thus 52.8 g of CO2 is produced.</u>

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The normal boiling point of iodomethane, CH3I, is 42.43 8C, and its vapor pressure at 0.00 8C is 140. Torr. Calculate (a) the st
tigry1 [53]

Answer:

a=28600J; b=90.6 J/K; c=402 torr

Explanation:

(a) considering the data given

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Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)

Using the Clausius-Clapeyron Equation

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(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.

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P298 K = 0.529 atm

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