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Goshia [24]
3 years ago
8

Combustion of 6.90g of this compound produced 13.8g of CO2 and 5.64g of H2O. What is the empirical formula of the unknown compou

nd?
Chemistry
1 answer:
Alla [95]3 years ago
4 0
The empirical formula for the unknown compound would be: C2H4O (2 molecules of Carbon, 4 molecules of Hydrogen, and 1 molecule of Oxygen)
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Strontium-90 is one of the products of the fission of uranium-235. This strontium isotope is radioactive, with a half-life of 28
Firlakuza [10]

Answer:

67.6 years is the time the isotope take to decay from 0.900g to 0.170g

Explanation:

The radioactive decay follows first order law:

Ln [A] = -kt + ln[A]₀

<em>Where [A] is concentration after time t,</em>

<em>k is decay constant:</em>

<em>k = ln 2 / t(1/2)</em>

<em>k = ln2 / 28.1 years</em>

<em>k = 0.02467 years⁻¹</em>

<em>[A]₀ = Initial concentration.</em>

<em />

We can replace concentration and use the mass of the isotope:

Ln [A] = -kt + ln[A]₀

Ln [0.170g] = -0.02467 years⁻¹t + ln[0.900g]

-1.667 = -0.02467 years⁻¹t

t =

<h3>67.6 years is the time the isotope take to decay from 0.900g to 0.170g</h3>
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Aonde a civilização egípcia se desenvolvera​
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Why are covalent substances gases and liquid rather than solids?
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Covalent compounds are held together with an intra molecular attraction which is weaker than metallic bond

hence covalent compounds exist as liquids, gases and soft solids

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Answer:

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