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lisabon 2012 [21]
2 years ago
8

7. A certain hydrocarbon, CxHy, is burned (reacts with O2 gas) and produces 1.955 g of CO2 for every

Chemistry
1 answer:
podryga [215]2 years ago
4 0

Answer:

The empirical formula of the hydrocarbon is C₂H₅

Explanation:

The formula for the hydrocarbon is C_xH_y

The mass of CO₂ produced per 1,000 g of H₂O = 1,955 g

The equation for the chemical reaction is given as follows;

C_xH_y + (x + y/4) O₂ → XCO₂ + y/2H₂O

From the given chemical equation, counting the number of atoms on both sides of the equation, we have;

The molar mass of CO₂ = 44.01 g/mol

The molar mass of H₂O = 18.01528 g/mol

The number of moles of H₂O in 1,000 g of H₂O = 1,000 g/(18.01528 g/mol) = 55.5084351 moles

The number of moles of CO₂ in 1,955 g of H₂O = 1,955 g/(44.01 g/mol) = 44.4217223 moles

Therefore, given that X moles of CO₂ is produced alongside Y/2 moles of H₂O. we have;

X = 44.4217223, Y/2 = 55.5084351

∴ Y = 2 × 55.5084351 = 111.0168702

The ratio of X to Y = X/Y = 44.4217223/111.0168702 = 0.40013488238

∴ The ratio of X to Y = X/Y ≈ 0.4 = 4/10

X/Y ≈ 4/10

The empirical formula is the representation of molecular formula in the smallest whole number ratio of the elements of the molecules

Therefore, when X = 4, Y = 10, from which we have the smallest ratio as;

When X = 2, Y = 5

The empirical formula of the hydrocarbon is therefore, C_xH_y = C₂H₅

The given chemical equation becomes;

C₂H₅ + (2 + 5/4) O₂ → 2CO₂ + 5/2H₂O

C₂H₅ + 3.25 O₂ → 2CO₂ + 2.5 H₂O

We then have;

4C₂H₅ + 13 O₂ → 8CO₂ + 10 H₂O

The empirical formula of the hydrocarbon, C_xH_y = C₂H₅.

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Which actions are part of a system of methodical tests and refinements specific to technological design? 1 documenting, tinkerin
telo118 [61]

Answer:

1. documenting, tinkering, testing

Explanation:

Technological design is defined as the process of study, design and development of new technologies.

There are some action in the methodical tests and refinements specific to technological design include documenting, tinkering, testing.

<u>Documenting </u><u>includes collecting all the information about the design and develop the product, </u><u>tinkering</u><u> involves repairing or adjust the issues found in the development, and </u><u>testing </u><u>helps to evaluate if the product is ready to work as it is supposed to.</u>

Hence, the correct answer is "1."

6 0
3 years ago
A- south<br> B- north <br> C-west<br> D- east <br> Help ASAP no links or I’m reporting
Agata [3.3K]

Answer:

east

Explanation:

5 0
3 years ago
What is the molarity of a nitric acid solution if 43.13 mL 0.1000 M KOH solution is needed to neutralize 30.00 mL of the acid so
Leno4ka [110]

Answer:

0.144M

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

HNO3 + KOH —> KNO3 + H20

From the equation,

nA = 1

nB = 1

From the question given, we obtained the following:

Ma =?

Va = 30.00mL

Mb = 0.1000M

Vb = 43.13 mL

MaVa / MbVb = nA/nB

Ma x 30 / 0.1 x 43.13 = 1

Cross multiply to express in linear form

Ma x 30 = 0.1 x 43.13

Divide both side by 30

Ma = (0.1 x 43.13) /30 = 0.144M

The molarity of the nitric acid is 0.144M

8 0
3 years ago
2. _____ is the distance and direction of an object’s change in position from the starting   a.  Speed b.  Displacement   c.  Ve
Llana [10]
Displacement is the distance and direction of an object's change in position from the starting.
Hence option B is correct.

Hope this helps!
3 0
3 years ago
The average atomic mass of Eu is 151.96 amu. There are only two naturally occurring isotopes of europium, Eu with a mass of 151.
earnstyle [38]

Answer:

The percentage abundance of Eu isotopes are 52 %  and 48 % .

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope,:

% = x %

Mass = 151.0 amu

For second isotope :

% = 100  - x  

Mass = 153.0 amu

Given, Average Mass = 151.96 amu

Thus,

151.96=\frac{x}{100}\times {151.0}+\frac{100-x}{100}\times {153.0}

Solving for x, we get that:

x = 52 %

<u>Thus percentage abundance of Eu isotopes are 52 %  and 48 % .</u>

7 0
3 years ago
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