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4 years ago

What is the density of a cube that has a mass of 25 g in a volume of 125 cm 3

1 answer:

3 0

<span>The density of an object is defined to be its mass divided by the volume it occupies. For this problem, the mass of the cube was given to be 25 g while its volume is 125 cm</span>³. Thus, we simply divide 25 g by 125 cm³ to get the object’s density. We then calculate that the cube has a density of 0.2 g/ cm³.

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Convert 682000000000 kmol to Tmol

Elena-2011 [213]

Answer:

682 teramoles

Explanation:

5 0

3 years ago

Read 2 more answers

The molar volume of oxygen,O2, is 3.90 dm3 mol-1 at 10.0 bar and 200 degree centigrade. Assuming that the expansion may be trunc

schepotkina [342]

Answer:

B = - 0.0326 dm³/mol

Explanation:

virial eq until second term:

PVm = RT [ 1 + B/Vm ]

∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm

∴ T = 200°C = 473 K

∴ Vm = 3.90 dm³/mol

∴ R = 0.08206 dm³.atm/K.mol

⇒ PVm / RT = 1 + B/Vm

⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm

⇒ 0.99164 = 1 + B/Vm

⇒ B/Vm = - 8.357 E-3

⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )

⇒ B = - 0.0326 dm³/mol

4 0

3 years ago

Word equation of almininum reaction with oxygen

zheka24 [161]

Answer:

Aluminium + oxygen = almininum oxide

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4 0

3 years ago

Why is argon monoatomic in nature

siniylev [52]

Argon monatomic in nature because it is a noble gas and has completely filled shells and thus does not form a bond with other elements.

7 0

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 $(\Delta S^o)$ .

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

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