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Nonamiya [84]
3 years ago
9

Scientists Pam and Alejandro were discussing a liquid found in the science lab that was missing it's label. In classifying the u

nknown liquid, they first needed to determine if the liquid is a mixture or pure substance. How is a pure substance different from a mixture?
Question 11 options:


A.Mixtures cannot be separated by physical meansA pure substance is heterogeneous.


B.Pure substances cannot be separated by physical means.

C.mixture is made of one substance.

D. a pure substance is heterogeneous.
Chemistry
2 answers:
strojnjashka [21]3 years ago
6 0

Your answer for this would be A. Mixtures cannot be separated by physical meansA pure substance is heterogeneous.


Hope This Helps :)

elena-14-01-66 [18.8K]3 years ago
3 0

Answer is: B. Pure substances cannot be separated by physical means.

Pure substance is made of only one type of atom (element) or only one type of molecule (compound), mixtures and solutions are made from two or more types of pure substances.

Elements (for example copper, iron, sulfur) and compounds (water, sodium chloride) have definite and constant composition with distinct chemical properties.  

Pure substances can be separated chemically, not physically, that is difference between pure substances and mixtures.

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Answer : The enthalpy of formation of H_2SO_4 is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of H_2SO_4 will be,

S+H_2+2O_2\rightarrow H_2SO_4    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) S+O_2\rightarrow SO_2     \Delta H_1=-298.2

(2) SO_2+\frac{1}{2}O_2\rightarrow SO_3    \Delta H_2=-98.2

(3) SO_3+H_2O\rightarrow H_2SO_4    \Delta H_3=-130.2

(4) H_2+\frac{1}{2}O_2\rightarrow H_2O    \Delta H_4=-285.8

Now adding all the equations, we get the expression for enthalpy of formation of H_2SO_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)

\Delta H=-812.4kJ/mol

Therefore, the enthalpy of formation of H_2SO_4 is, -812.4 kJ/mole

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