Chemistry, 26.06.2019 08:30 Lesquirrel

Which of the following statements does not accurately describe precambrian earth? a. the atmosphere was similar to that of early mars and venus. b. the oceans were chemically different from modern oceans. c. the ocean water was salty and rich in dissolved oxygen. d. there were large amounts of water in earth’s mantle.

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Answer from: joThompson

C. The ocean water was salty and rich in dissolved oxygen.

Explanation:

The oldest and longest of the Geological Ages, the Pre-Cambrian extends from the formation of the Earth, approximately 4.5 billion years ago, to 570 million years ago.

This period begins with the formation of the Earth about 4.5 billion years ago, where the Earth was an incandescent mass with rivers of dissolved rock, erupting volcanoes and large amount of sulfur. The atmosphere was initially similar to the atmosphere of Mars and Venus and was composed of nitrogen, ammonia, hydrogen, carbon monoxide, methane and water vapor, expelled by volcanoes, and no sign of oxygen.

After 700 million years, most of the rocks on the surface of the earth had already been cooled and a large amount of water had condensed into an ocean. Even so, the volcanic eruptions were still constant and were giving rise to several islands that were pushed toward each other by the movement of the mantle. In this period small unicellular beings in the ocean begin to appear, the primitive algae and bacteria that assimilate the carbon dioxide of the atmosphere transforming it into free oxygen, slowly changing the composition of the atmosphere that, in that period, was composed basically by nitrogen and steam of Water. They are of this period the oldest fossils of the earth.

Answer from: Quest

a compound is a molecule made of atoms from different elements.

a. by the combination of an unlimited number of elements

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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