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3 years ago

Measurements show that unknown compound X has the following composition:

Chemistry

2 answers:

lozanna [386]3 years ago

8 0

the empirical chemical formula of X is CH4

Explanation:Step 1: Imagine you have a sample of compound weighing exactly . Multiply the mass of this sample by the mass percents to find the mass of each element in the sample

Step 2: Divide the mass of each element by the element's molar mass to find the moles of each element in the sample. Remember to round your answers to the correct number of significant digits.

Step 3: Divide the moles of each element by the the smallest number of moles of any element to find the mole ratio of elements in the sample.

Step 4: Multiply the mole ratio by the smallest whole number that changes it into a whole number ratio to find the atom ratio of elements in the sample.

Note that the result of each multiplication must equal a whole number only within measurement uncertainty.

The measurement uncertainty in this calculation comes from the measurement uncertainty of the mass percents given in the question. The mass percents each have significant digits. That means each mass percent has some measurement uncertainty in the third significant digit, and only the first two significant digits can be considered completely reliable.

In Step 1 you multiplied each mass percent by something with zero uncertainty (the exactly you assumed your sample weighed), in Step 2 you divided by a measurement with more than significant digits (the molar mass of the elements), and in Step 3 you divided by a measurement with the same number of significant digits (the least number of moles of any element in the compound). None of these steps added to the measurement uncertainty of your calculation.

Therefore, the measurement uncertainty in the final result of all your calculations is determined by the measurement uncertainty in the original mass percents and will be in the third significant digit. That means the result of each of the final multiplication steps must equal a whole number only to within the first two significant digits.

The whole numbers in the last column of the table are and .

Dominik [7]3 years ago

4 0

X is carbon and Jess

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3 years ago

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Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

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3 years ago

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Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0

olya-2409 [2.1K]

Answer:

$pH=-1.37$

Explanation:

We are given that 25 mL of 0.10 M $CH_3COOH$ is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of $CH_3COOH=25mL=0.025 L$

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL= $\frac{35}{1000}=0.035 L$

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of $CH_3COOH$ =0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+= $0.10\times 0.025$ =0.0025 moles

Number of moles of $OH^-=0.10\times 0.01$ =0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+= $\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L$

pH=-log [H+]=-log [4.28 $\times 10^{-2}$ ]=-log4.28+2 log 10=-0.631+2

$pH=-1.37$

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The answer to the question

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