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Elza [17]
3 years ago
15

When rubidium metal is exposed to air, two atoms of rubidium, Rb, combine with one atom

Chemistry
1 answer:
Eddi Din [679]3 years ago
8 0

2.1653 g

Explanation:

The molar mass of Rubidium is;

85.468 g/mol

Therefore the moles of Rubidium that reacted with oxygen is;

1.98 / 85.468

= 0.0232 moles

If every two moles of Rubidium reacts with one mole of oxygen then the amount of oxygen consumed in the chemical reaction is;

0.5 * 0.0232

= 0.0116 moles

The molar mass of an oxygen atom is 16 g/mole. Then the amount of O in grams consumed is;

0.0116 * 16

=0.1853 g

The final weight of the Rubidium II Oxide is;

1.98 + 0.1853

= 2.1653 g

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Answer:

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In 1st equilibrium, H₃O⁺ is the conjugate acid and HCO₃⁻ the conjugate base.

In 2nd equilibrium, NH₄⁺ is the conjugate acid, and OH⁻, the conjugate base.

Explanation:

By the Bronsted-Lowry you know that acids are the one that release protons and base are the ones that catch them.

For the first equilibrium:

H₂CO₃(aq) + H₂O(l) ⇄ H₃O⁺(aq) + HCO₃⁻(aq)

Carbonic acid is the acid → It donates the proton to water, so the water becomes the base. As H₂CO₃ is the acid,  the bicarbonate is the conjugate base (it can accept the proton from water to become carbonic acid, again) and the hydronium is the conjugate acid (it would release the proton to become water).

For the second equilibrium:

NH₃(aq) + H₂O(l) ⇄  NH₄⁺ (aq) + OH⁻(aq)

This is the opposite situation → Water relase the proton to ammonia, that's why water is the acid and NH₃, the base (it accepted to become ammonium). The NH₄⁺ is the conjugate acid (it can release the H⁺ to become ammonia) and the OH⁻ is the conjugate base (It can accept the proton to become water, again).  

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Therefore, the correct answer is 0.450 M H2SO4. Note that the answer has 3 significant figures because each of the given values also contains 3 significant figures.

Hope this helps!

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