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ddd [48]
3 years ago
5

What is the empirical formula of the compound with the molecular formula c18h36?

Chemistry
1 answer:
emmasim [6.3K]3 years ago
6 0
When we write a formula of a compound in its simplest form, it is empirical formula.
Given the molecular formula is = C₁₈H₃₆
Now we can relate both empirical and molecular formula as;
Molecular formula = n x empirical formula
C₁₈H₃₆ = 18 x CH₂
So the empirical formula of C₁₈H₃₆ is CH₂
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Why is an atom regarded as electrically unstable​
meriva

Answer: An atom can be considered unstable in one of two ways. If it picks up or loses an electron, it becomes electrically charged and highly reactive. Such electrically charged atoms are known as ions. Instability can also occur in the nucleus when the number of protons and neutrons is unbalanced.

Explanation:

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Twenty irregular pieces of an unknown metal have a collective mass of 28.225 g. When carefully placed in a graduated cylinder th
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Explanation:

Use the density formula to determine the volume of the piece of metal.

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8 0
4 years ago
What is the value of the equilibrium constant at 25 oC for the reaction between the pair: I2(s) and Br-(aq) Give your answer usi
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Explanation:

Formula to calculate standard electrode potential is as follows.

          E^{o}_{cell} = E^{0}_{cathode} - E^{0}_{anode}

                             = 0.535 - 1.065

                             = - 0.53 V

Also, it is known that relation between E^{o}_{cell} and K is as follows.

            E^{o}_{cell} = \frac{RT}{nF} \times ln K

                 ln K = \frac{nFE^{0}_{cell}}{RT}      

Substituting the given values into the above formula as follows.

                 ln K = \frac{nFE^{0}_{cell}}{RT}    

                        =  \frac{2 \times 96485 C mol^{-1} \times -0.53 V}{8.314 l atm/mol K \times 298 K} \times \frac{1 J}{1 V C}  

                ln K = -41.28

                    K = e^{-41.28}    

                        = 1 \times 10^{-18}

Thus, we can conclude that the value of the equilibrium constant for the given reaction is 1 \times 10^{-18}.      

5 0
3 years ago
A student calculated the density of a sample of graphite to be 2.3 g/cm3. Show a numerical setup for calculating the student’s p
LUCKY_DIMON [66]

Answer:

Percent error = 1.5%

Explanation:

Given data:

Measured value of density of graphite = 2.3 g/cm³

Percent error = ?

Solution:

Formula:

Percent error = [Measured value - Actual value / actual value] × 100

Actual/accepted value of density of graphite = 2.266 g/cm³

Now we will put the values:

Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100

Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100

Percent error = 0.015  × 100

Percent error = 1.5%

8 0
3 years ago
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A. As pressure on the gas increases, the volume and temperature will both decrease

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