What do the water, nitrogen, and carbon cycle all have in common

Chemistry

1 answer:

Lelu [443]3 years ago

5 0

Answer:

How are the water carbon and nitrogen cycles related?

The three main cycles of an ecosystem are the water cycle, the carbon cycle and the nitrogen cycle. These three cycles working in balance are responsible for carrying away waste materials and replenishing the ecosystem with the nutrients necessary to sustain life.

Explanation:

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This is the balances equations. C3H8 + 5O2 → 3CO2 + 4H2O How many moles of oxygen are required to produce 37.15 g CO2

AleksandrR [38]

Molar mass O2 = 31.99 g/mol

Molar mass CO2 = 44.01 g/mol

Moles ratio:

<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2

mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01

mass of O2 = 8174.8575 / 132.03

mass of O2 = 61.916 g

Therefore:

1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916

moles O2 = 61.916 x 1 / 31.99

moles = 61.916 / 31.99 => 1.935 moles of O2

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3 years ago

What is meant by localised orbitals? Explain.

Anni [7]

Localized molecular orbitals are molecular orbitals which are concentrated in a limited spatial region of a molecule, for example a specific bond or a lone lake on a specific atom.

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3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$