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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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A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this p
xxTIMURxx [149]

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

Moles=1.20 \times {25.0\times 10^{-3}}\ moles

<u>Moles of potassium chloride  = 0.03 moles</u>

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles=0.900 \times {15.0\times 10^{-3}}\ moles

<u>Moles of lead(II) nitrate  = 0.0135 moles</u>

According to the given reaction:

2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

<u>Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

<u>Theoretical yield = 3.75435 g</u>

Given experimental yield = 2.45 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %</u>

6 0
2 years ago
List the following compounds in decreasing electronegativity difference. cl2 hcl nacl
Mkey [24]
Based on Pauling Scale, electro negativity of Cl = 3.2, Na = 0.9 and H = 2.1

Thus, Electronegativity difference  in Cl_{2} = 3.2 -3.2 = 0
Electronegativity difference  in NaCl = 3.2-0.9 = 2.3
Similarly, Electronegativity difference  in HCl = 3.2 - 2.1 = 1.1

Thus, among the listed molecules following is the decreasing order of electronegativity difference: NaCl> HCl > Cl_{2}
8 0
2 years ago
An ideal gas occupies a volume V at an absolute temperature T. If the volume is halved and the pressure kept constant, what will
Kruka [31]

Answer:

It will be halve of T

Explanation:

V1 = V

T1 = T

V2 = ½V

T2 = x

V1/T1 = V2/T2

V/T = ½V/x

Vx = ½VT

2Vx = VT

2x = T

x = ½T

6 0
2 years ago
4. The stockroom contains 1.0 M NaAc (Sodium Acetate), 1.0 M HAc (acetic Acid), distilled water and strong acids and bases. You
Elina [12.6K]

Answer:

Concentrations of HAc and NaAc you need are 0.122M

Explanation:

pKa of acetic acid is 4.75, that means when amount of sodium acetate and acetic acid is the same, pH will be 4.75

Thus, you know [NaAc]i = [HAc]i

Now, using H-H equation, when pH = 3.75:

3.75 = 4.75 + log [NaAc] / [HAc]

0.1 = [NaAc] / [HAc]

10 [NaAc] = [HAc]

Thus, after the reaction  [HAc] must be ten times,  [NaAc].

Based in the reaction of NaAc with HCl

NaAc + HCl → HAc + NaCl

Moles of HCl added are:

1mL = 0.001L * (10mol /L) = 0.01 moles HCl.

That means moles of both compounds after the reaction are:

<em>[NaAc] = [NaAc]i - 0.01 mol </em>

[HAc] = [HAc]i + 0.01

Replacing these equations with the information you know:

[NaAc] = [NaAc]i - 0.01 mol

10[NaAc] = [NaAc]i + 0.01

Subtracting both equations:

9[NaAc] = 0.02mol

[NaAc] = 0.0022 moles.

Replacing in <em>[NaAc] = [NaAc]i - 0.01 mol </em>

0.0022mol = [NaAc]i - 0.01 mol

0.0122mol = [NaAc]i = [HAc]i

These moles in 100.00mL = 0.1000L:

[NaAc]i = [HAc]i = 0.0122mol / 0.100L =

0.122M

Thus, <em>concentrations of HAc and NaAc you need are 0.122M</em>

<em />

To create this buffer, you need to pipette 12.2mL of both 1.0M NaAc and 1.0M HAc and dilute this mixture to 100.0mL

4 0
2 years ago
Make a quantitative and qualitative statement about the following:. water, carbon, iron, hydrogen gas, sucrose, table salt, merc
gladu [14]
The elements in this list are mercury, gold, iron, carbon and hydrogen. The compounds in this list, on the other hand, are sucrose, table salt, water and air. Elements are composed only of one substance while compounds are composed of two or more substances.
8 0
2 years ago
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