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2 years ago

(Gamergirl223)

Chemistry

1 answer:

34kurt2 years ago

3 0

The answer is: To demonstrate that the liquid is a mixture, the scientist can perform the chromatography.

Mixture: They are formed when two elements or compounds are combined physically and not chemically.

What are the types of mixtures?

Mixtures are divided into two categories :Homogeneous and heterogeneous.

Homogeneous mixtures- They have the same composition throughout their mass and there are no visible boundaries between the components of the mixture. Example- Sugar solution, salt solution, alloys, alcohol dissolved in water.

Heterogeneous mixture- They do not have the same composition throughout there are visible boundaries between the components of the mixture. Example- Milk, sand, and iron fillings, solution of $CaCO_3$ , solution of $K_2Cr_2O_7$ .

To demonstrate the the given liquid is a mixture, the scientist can perform Chromatography.
Chromatography is based on the different rates of adsorption of components of a mixture on a suitable adsorbent.
When a drop of a mixture of substances is put on a chromatogram paper and the paper is dipped in a solvent, the substance that is more soluble in the solvent rises faster compared to others. And due to this difference in adsorption of different substances, they are separated by the method of chromatography.
If two spots are seen on a chromatogram that means the liquid is a mixture and if only one spot is visible, that suggests that liquid is a pure liquid.

To learn more about the mixture and chromatography, visit:

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Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe

ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

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.

(a) The formation of $CH_4$ will be,

$2C(coal)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(coal)+H_2O(g)\rightarrow CO(g)+H_2(g)$ $\Delta H_1=29.7kJ$

(2) $CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)$ $\Delta H_2=-41kJ$

(3) $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ $\Delta H_3=-206kJ$

We are multiplying equation 1 by 2 and then adding all the equations, we get :

(b) The expression for enthalpy of reaction will be,

$\Delta H_{rxn}=2\times \Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(2\times 29.7)+(-41)+(-206)$

$\Delta H_{rxn}=-187.6kJ/mol$

Therefore, the enthalpy of reaction is, -187.6 kJ/mol

(c) Now we have to calculate the total heat.

$\Delta H=\frac{q}{n}$

or,

$q=\Delta H\times n$

where,

$\Delta H$ = enthalpy change = -187.6 kJ/mol

q = heat = ?

n = number of moles of coal = $\frac{1.00\times 1000g}{12.00g/mol}=83.33mol$

Now put all the given values in the above formula, we get:

$q=(-187.6kJ/mol)\times (83.33mol)=-2.251kJ$

Thus, the total heat will be, -2251 kJ

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4 years ago

The free energy of formation of nitric oxide, NO, at 1000 K (roughly the temperature in an automobile engine during ignition) is

Assoli18 [71]

<u>Answer:</u> The value of $K_p$ for the chemical equation is $8.341\times 10^{-5}$

<u>Explanation:</u>

For the given chemical equation:

$N_2(g)+O_2(g)\rightarrow 2NO(g)$

To calculate the $K_p$ for given value of Gibbs free energy, we use the relation:

$\Delta G=-RT\ln K_p$

where,

$\Delta G$ = Gibbs free energy = 78 kJ/mol = 78000 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = 1000 K

$K_p$ = equilibrium constant in terms of partial pressure = ?

Putting values in above equation, we get:

$78000J/mol=-(8.314J/Kmol)\times 1000K\times \ln K_p\\\\Kp=8.341\times 10^{-5}$

Hence, the value of $K_p$ for the chemical equation is $8.341\times 10^{-5}$

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4 years ago

How many moles of carbon dioxide are produced when 5.12 mol of propane gas is burned in excess oxygen? C3H8(g) + 5O2(g) → 3CO2(g

maxonik [38]

Answer:

15.4 mol CO₂

Explanation:

5.12 mol C₃H₈ × (3 mol CO₂ / mol C₃H₈) ≈ 15.4 mol CO₂

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bogdanovich [222]

Explanation:

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4 years ago

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