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3 years ago

How is a pure substance different from other substances?

Chemistry

1 answer:

iren [92.7K]3 years ago

7 0

Answer:

Explanation:

A pure substance is made up solely of that substance and can't be separated into any other substances. A mixture can be separated into two or more pure substances

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The volume of the earth is approximately 1.08326x10 to the 12th power km to the 3rd power. How many liters is this

mafiozo [28]

Here, we are required to determine the volume of the earth which is 1.08326 × 10¹² km³ in liters.

The volume of the earth is approximately,

, 1.08326 × 10²⁴ liters

By conversion factors;

1dm³ = 1liter
However; 1km = 10000dm = 10⁴ dm
Therefore, 1km³ = (10⁴)³ dm³.

Consequently, 1km³ = 10¹²dm³ = 10¹²liters.

The conversion factor from 1km³ to liters is therefore, c.f = 10¹²liters/km³

Therefore, the volume of the earth which is approximately, 1.08326 × 10¹² km³ can be expressed in liters as;

1.08326 × 10¹² km³ × 10¹²liters/km³

The volume of the earth is approximately,

1.08326 × 10²⁴ liters.

Read more:

brainly.com/question/16814684

7 0

2 years ago

The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc

jek_recluse [69]

Answer: $K_{sp}=1.25\times 10^{-14}$

$pK_{sp}=13.90$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of magnesium phosphate is given as:

$Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}$

When the solubility of $Mg_3(PO_4)_2$ is S moles/liter, then the solubility of $Mg^{2+}$ will be 3S moles\liter and solubility of $PO_4^{3-}$ will be 2S moles/liter.

Thus S = 0.173 g/L or $\frac{0.173g/L}{262.8g/mol}=0.00065mol/L$

$K_{sp}=(3S)^3\times (2S)^2$

$K_{sp}=108S^5$

$K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}$

$pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90$

5 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 $(\Delta S^o)$ .

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

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