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Marrrta [24]
4 years ago
9

The oxygen revolution changed Earth’s environment dramatically. Which of the following adaptations took advantage of the presenc

e of free oxygen in the oceans and atmosphere? Select one of the options below as your answer:
A. The evolution of cellular respiration, which used oxygen to help harvest energy from organic molecules.
B. The evolution of chloroplasts.
C. The persistence of some animal groups in anaerobic habitats.
Chemistry
1 answer:
denpristay [2]4 years ago
4 0
The correct answer is A. Oxygen became used to carry out many processes and to drive a few anaerobic processes.
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How do I go from molecules to moles? For example 1.50e23 molecules NH3 how many moles are in this problem?
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Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti
ella [17]

Answer :

The mass of excess mass of Na_2CO_3, AgNO_3,Ag_2CO_3\text{ and }NaNO_3 are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

Mass of Na_2CO_3 = 4.25 g

Mass of AgNO_3 = 7.50 g

Molar mass of Na_2CO_3 = 106 g/mole

Molar mass of AgNO_3 = 170 g/mole

Molar mass of Ag_2CO_3 = 276 g/mole

Molar mass of NaNO_3 = 85 g/mole

First we have to calculate the moles of Na_2CO_3 and AgNO_3.

\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles

\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3

From the balanced reaction we conclude that

As, 2 moles of AgNO_3 react with 1 mole of Na_2CO_3

So, 0.044 moles of AgNO_3 react with \frac{0.044}{2}=0.022 moles of Na_2CO_3

From this we conclude that, Na_2CO_3 is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

The excess mole of Na_2CO_3 = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of Na_2CO_3.

\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g

Now we have to calculate the moles of Ag_2CO_3.

As, 1 moles of AgNO_3 react to give 1 moles of Ag_2CO_3

So, 0.044 moles of AgNO_3 react to give 0.044 moles of Ag_2CO_3

Now we have to calculate the mass of AgCO_3.

\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g

Now we have to calculate the moles of NaNO_3.

As, 2 moles of AgNO_3 react to give 2 moles of NaNO_3

So, 0.044 moles of AgNO_3 react to give 0.044 moles of NaNO_3

Now we have to calculate the mass of NaNO_3.

\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g

6 0
4 years ago
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