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3 years ago

How many molecules are there in 8.00 grams of glucose, C6H12O6

Chemistry

1 answer:

beks73 [17]3 years ago

6 0

Answer:

molecules C6H12O6 = 2.674 E22 molecules.

Explanation:

from periodic table:

⇒ molecular mass C6H12O6 = ((6)(12.011)) + ((12)(1.008)) + ((6)(15.999))

⇒ Mw C6H12O6 = 180.156 g/mol

⇒ mol C6H12O6 = (8.00 g)(mol/180.156 g) = 0.0444 mol C6H12O6

∴ mol ≡ 6.022 E23 molecules

⇒ molecules C6H12O6 = (0.0444 mol)(6.022 E23 molecules/mol)

⇒ molecules C6H12O6 = 2.674 E22 molecules

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2 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 $(\Delta S^o)$ .

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

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