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Crank
3 years ago
12

Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h

ydrocarbon?
Show work please?
Chemistry
1 answer:
cluponka [151]3 years ago
3 0
<span>C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5</span>
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Which of the following observations about pH is consistent for an acidic solution?
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A pH value of 3

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Explain the process of groundwater
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Limestone is easily eroded from above and below.

Explanation:

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3 0
3 years ago
The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1
Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

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3 0
3 years ago
In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated so
gulaghasi [49]

<u>Answer:</u> The moles of hydroxide ions present in the sample is 0.0008 moles

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl.

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL

Putting values in above equation, we get:

1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles

8 0
3 years ago
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