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3 years ago

What can not be separated easily or, sometimes at all

1 answer:

8 0

Pure Substances cannot be separated easily or, sometimes at all.

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An empty steel container is filled with 2.0 atm of H₂ and 1.0 atm of F₂. The system is allowed to reach equilibrium according to

Darina [25.2K]

H₂ (g) + F₂ (g) ⇌ 2 HF (g)

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3 years ago

What Went Wrong – Balancing Chemical Equations

OLEGan [10]

1-It has to be 3 Fe and not Fe3.
2-The oxygens aren't balanced

Balanced equation:
3Fe+4H2O---->Fe3O4+4H2

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3 years ago

Why do foods and fuels have chemical potential energy?

Ainat [17]

Fuels are burned to produce energy, and your body breaks down the food to produce nutrients for your body.

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3 years ago

Read 2 more answers

An impurity sometimes found in Ca₃(PO₄)₂ is Fe₂O₃, which is removed during the production of phosphorus as ferrophosphorus (Fe₂P

Citrus2011 [14]

This impurity is troubling from an economic standpoint because it lead to decrease in the yield of phosphorus

Ferrophosphorus is a byproduct of phosphorus production in submerged-arc furnaces , by their reduction with carbon. It is formed from the iron oxide impurities.

Iron impurities present in the calcium phosphate will be precipitated out as the iron phosphate which eventually will lead to the decrease in the yield of phosphorous during the production of phosphorous.

Thus we can conclude that Fe₂P causes decrease in yield

Learn more about production of phosphorus at brainly.com/question/13337198

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1 year ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

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3 years ago