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Novay_Z [31]
2 years ago
9

Complete combustion of 1.00 g of the hydrocarbon pagodane gives 3.38 g carbon dioxide. what is the empirical formula of pagodane

?
Chemistry
1 answer:
Blababa [14]2 years ago
6 0
Combustion of any hrdrocarbon yields carbon dioxide and water such that the hydrogen and carbon are derived from the hydrocarbon. 
1 mole of carbon dioxide has a mass out of which 12 g is the mass of carbon, therefore, 3.38 g of carbon will contain  (3.38 × 12)/44 = 0.922 g of carbon.
Thus, since the hydrocarbon has a mass of 1 g, then the mass of hydrogen will be (1-0.922 g) = 0.078 g.
To get the empirical formula we divide the mass of each element by the atomic mass to get the number of moles.
Carbon= 0.922/12 = 0.0768 moles
Hydrogen = 0.078/1 = 0.078 moles
Then we get the ratio of the moles of carbon : hydrogen
                                                      =   0.0768 : 0.078
                                                         =         1 : 1.016
                                                          ≈        1:   1
Therefore the empirical formula of the hydrocarbon will be CH


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How much heat is released during the formation of 3.18 mol HCl(g) in this reaction: H2(g)+Cl2(g) → 2HCl(g) with a H of -184.6 kJ
Vikentia [17]

The given thermochemical reaction is between hydrogen gas and chlorine gas to form hydrogen chloride.

This can be represented as:

H_{2}(g)+Cl_{2}(g)-->2HCl(g)  ΔH_{reaction}^{0}=-184.6 kJ/mol

So when two moles of HCl is formed, 184.6 kJ of energy is released.

Calculating the heat released when 3.18 mol HCl (g) is formed in the reaction:

3.18 molHCl*\frac{184.6kJ}{2molHCl} =293.5 kJ

Therefore, 293.5 kJ of heat is released when 3.18 mol HCl is formed in the reaction between hydrogen and chlorine.



4 0
2 years ago
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Write a net ionic equation for the reaction that occurs when nickel(ii carbonate and excess hydrobromic acid (aq are combined.
seropon [69]
First, we write the reaction equation:
NiCO₃ + 2HBr → NiBr₂ + H₂CO₃

Now, writing this in ionic form:
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Overall equation:
NiCO₃ + 2Br⁻ → NiBr₂ + CO₃⁻²
7 0
1 year ago
Compute the end-to-end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule with a number-
MArishka [77]

Answer:

5156  Â

Explanation:

Chains of monomers that are being linked together constitute what is called polymers. in this question; we are to compute the end to end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule.

Using the following approach:

The first step is to determine the repeating units in polymer

The number of repeating unit = average molecular weight/  atomic weight of PVC

Given that:

average molecular weight = 256,131 g/mol

atomic weight of PVC = 62.498 g

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The number of repeating unit = 256,131 g/mol /  62.498 g = 4098.227143 mol

Now the distance is calculated by using the formula:

d = (C-C) × sin 109.5/2

C-C bond length = 1.54 angstroms

tetrahedral bond angle = 109.5°

Then d = (1.54)  × sin 109.5/2

d = 1.258

Thus ; the end to end separation  is :

4098.227143  × 1.258 = 5155.57 Â

The answer is 5156  Â (since no decimal should be included )

6 0
1 year ago
A student dissolved a sample in hexane, spotted it on to a TLC plate and eluted using ethyl acetate. After visualizing the TLC p
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Back to TLC

Original content © University of Colorado at Boulder, Department of Chemistry and Biochemistry.
The information on these pages is available for academic use without restriction.
3 0
1 year ago
How many grams of NO are required to produce 145 g of N2 in the following reaction?
V125BC [204]

Answer:

b. 186 g

Explanation:

Step 1: Write the balanced equation.

4 NH₃(g) + 6 NO(g) → 5 N₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 145 g of N₂

The molar mass of nitrogen is 28.01 g/mol.

145g \times \frac{1mol}{28.01 g} =5.18 mol

Step 3: Calculate the moles of NO required to produce 5.18 moles of N₂

The molar ratio of NO to N₂ is 6:5.

5.18molN_2 \times \frac{6molNO}{5molN_2} = 6.22molNO

Step 4: Calculate the mass corresponding to 6.22 moles of NO

The molar mass of NO is 30.01 g/mol.

6.22mol \times \frac{30.01g}{mol} =186 g

4 0
1 year ago
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