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Paladinen [302]
3 years ago
15

Cosmone has the molecular formula C15H26O. Calculate the number of moles of carbon in a 8.35-mol sample of cosmone. Calculate th

e mass of a sample of cosmone that contains 6.58 x 1019atoms of hydrogen. Calculate the number of atoms of carbon m 8.00-g of cosmone.
Chemistry
1 answer:
bulgar [2K]3 years ago
8 0

Answer:

1) 125.25 moles of carbon

2) 9.35 × 10⁻⁴ grams of cosmone

3) 1.44×10²¹ atoms of carbon

Explanation:

1) The given compound, cosmone, with molecular formula C₁₅H₂₆O has 15 toms f carbon per molecule, therefore, we have

8.35 mole of C₁₅H₂₆O, will contain 15 × 8.35 moles of carbon which is 125.25 moles of carbon

2) The number of moles of hydrogen in 6.58 × 10¹⁹ atoms of hydrogen = 6.58 × 10¹⁹/(6.02 × 10²³) = 1.09 × 10⁻⁴ moles

1 mole of cosmone contains 26 moles of hydrogen, therefore;

1 mole of hydrogen will be contained in 1/26 moles of cosmone

1.09 × 10⁻⁴ moles of hydrogen will be contained in 1/26 × 1.09 × 10⁻⁴moles of cosmone = 4.204 × 10⁻⁶ moles of cosmone

The molar mass of cosmone = 15*12.0107 + 26*1.00784 + 1 * 15.999 = 222.36334 g/mol

Therefore, the mass of 4.204 × 10⁻⁶ moles of cosmone = 4.204 × 10⁻⁶×222.36334  = 9.35 × 10⁻⁴ grams of cosmone

3) The number of moles in 8.00 g of cosmone = 8/222.36 = 0.036 moles

Therefore, number of moles of carbon atoms in 0.036 moles of cosmone = 0.036/15 moles which is 0.0024 moles of carbon;

1 mole of carbon atoms contains 6.02 × 10²³ atoms. therefore;

0.0024 moles of carbon will contain 6.02 × 10²³ × 0.0024 carbon atoms which is 1.44×10²¹ atoms of carbon.

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Consider a solution containing .100 M fluoride ions and .126M hydrogen fluoride. The concentration of fluoride ions after the ad
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Given:

Concentration of Fluoride ions = 0.100 M
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Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution

Assume: 50:50 ratio of fluoride ions and HF

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Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

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