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bagirrra123 [75]
3 years ago
6

Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by the element. rb

Chemistry
1 answer:
devlian [24]3 years ago
3 0

Actually Rb or Rubidium in zero state has the following electron configuration:

<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>

 

However we can see that the ion has a 1 positive charge, which means that it lacks 1 electron, therefore the answer from the choices is:

<span>d. rb+: 1s22s22p63s23p64s23d104p6</span>
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Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3
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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:

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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

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2 years ago
A 8 kg cat is running 4 m/s. How much kinetic energy does it have
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Kinetic energy would equal 16 J 
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