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xxTIMURxx [149]
3 years ago
8

Can a physical change, change what a substance is? Exp

Chemistry
1 answer:
Margarita [4]3 years ago
7 0
Physical changes<span> are </span>changes<span> affecting the form of a chemical </span>substance<span>, but not its chemical composition. </span>Physical changes<span> are used to separate mixtures into their component compounds, but </span>can<span> not usually be used to separate compounds into chemical elements or simpler compounds.</span>
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c, 1A hope this helps


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Some species are no longer living but we know they existed one time. we know this because....
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Which solution whose POH is 12
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d

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2 years ago
What is the number of Co2 in a 220 gram
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5 0
3 years ago
Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3
Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving  chemical system, in which a reversible chemical reaction takes place.  When this situation is reached, it is observed that the  concentrations of substances, both reagents and reaction products,  they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations  and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }

You know:

  • Kc= 14.5
  • [H₂]= 0.322 M
  • [CH₃OH] =1.56 M

Replacing:

14.5=\frac{1.56}{[CO]*0.322^{2} }

Solving:

[CO]=\frac{1.56}{14.5*0.322^{2} }

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

8 0
3 years ago
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