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Serga [27]
3 years ago
5

An unknown compound contains only C , H , and O . Combustion of 5.90 g of this compound produced 11.8 g CO2 and 4.83 g H2O . Wha

t is the empirical formula of the unknown compound
Chemistry
1 answer:
Mamont248 [21]3 years ago
7 0

Answer:

C₂H₄O

Explanation:

In a compound that contains Cabon, hydrogen and oxygen, the combustion produce CO₂ from the carbon, and H₂O from the hydrogens. Using the mass of the products we can solve the moles of Carbon and hydrogen. The empirical formula is the simplest whole-number of atoms present in a molecule.

<em>Moles CO₂ = Moles C:</em>

11.8g CO₂ * (1mol / 44g) = 0.268 moles CO₂ = 0.268 moles C * (12g/mol) =

3.216g C

<em>Moles H₂O = 1/2 moles H:</em>

4.83g H₂O * (1mol / 18g) = 0.268 moles H₂O * (2 mol H / 1 mol H₂O) =

0.537 mol H * (1g/mol) = 0.537g H

<em>Mass O to find moles O:</em>

5.90g Sample - 3.216g C - 0.537g H = 2.147g O * (1mol / 16g) = 0.134 moles O

<em>Ratio of atoms -Dividing in 0.134 moles-:</em>

C = 0.268mol C / 0.134 mol O = 2

H = 0.537mol H / 0.134 mol O = 4

O = 0.134mol O / 0.134 mol O = 1

Empirical formula is:

<h3>C₂H₄O</h3>

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You place air in a sealed can at standard temperature and pressure (STP). You double its absolute temperature (K) while leaving
AnnZ [28]

Answer:

Pressure will double

Explanation:

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=> P₁/T₁ = P₂/T₂ => P₂ = P₁T₂/T₁

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P₂ = P₁(2T₁)/T₁ = 2P₁

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3 years ago
What type of specialized cell in the eye is used for detecting low levels of light?
Vladimir [108]

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Cone Cell

Explanation:

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3 years ago
For the reaction of hydrogen with iodine
fenix001 [56]

Answer:

r_{H_2} = \frac{-1}{2} r_{HI}

Explanation:

Hello!

In this case, considering the given chemical reaction:

H_2(g) + I_2(g) \rightarrow 2HI(g)

Thus, by applying the law of rate proportions, we can write:

\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

r_{H_2} = \frac{-1}{2} r_{HI}

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A sample of paper from an ancient scroll was found to contain 39.5% 14C content as compared to a present-day sample. The t1/2 fo
nirvana33 [79]

Answer

7665 years

Procedure

Let N₀ be the amount of carbon-14 present in a living organism. According to the radioactive decay law, the number of carbon-14 atoms, N, left in a dead tissue sample after a certain time, t, is given by the exponential equation:

N = N₀e^(-λt)

where λ is the decay constant which is related to half-life (T1/2) by the equation:

\lambda=\frac{ln(2)}{t_{\frac{1}{2}}}

Here, ln(2) is the natural logarithm of 2.

The percent of carbon-14 remaining after time t is given by N/N₀.

Using the first equation, we can determine λt.

The half-life of carbon-14 is 5,720 years, thus, we can calculate λ using the second equation, and then find t.

\lambda=\frac{ln(2)}{5720}=1.211\times10^{-4}

Solving the second equation for t, and using the λ we have just calculated we will have

t= 7665 years

3 0
1 year ago
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