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Zinaida [17]
3 years ago
8

Consider a sample of 10.0 g of the gaseous hydrocarbon C2H6 to answer the following question: How many moles are present in this

sample?
When answering the question, include the following:

State how to find the molar mass for the hydrocarbon.
State how you know if you need to multiply or divide by the molar mass.
Give the correct number of significant figures and explain why the answer has that many significant figures.
Chemistry
1 answer:
Masja [62]3 years ago
7 0

<u>Answer:</u> The moles of given hydrocarbon is 0.3 moles

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of ethane = 10.0 g

Molar mass of ethane = [(2\times 12)+(6\times 1)]=30g/mol

We need to divide the given value by the molar mass.

Putting values in above equation, we get:

\text{Moles of ethane}=\frac{10.0g}{30g/mol}=0.3mol

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits. Here, the least precise number of significant digits are 1.

Hence, the moles of given hydrocarbon is 0.3 moles

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Answer:

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7 0
3 years ago
The K sp for barium fluoride, BaF2, is 2.45 × 10-5. What is the molar solubility of barium fluoride?
lilavasa [31]

Answer: The molar solubility of barium fluoride is 0.0183 moles/liter.

Explanation:

The equation for the reaction will be as follows:

BaF_2\rightarrow Ba^{2+}+2F^

By Stoichiometry,

1 mole of BaF_2 gives 2 moles of F^- and 1 mole of Ba^{2+}

Thus if solubility of BaF_2 is s moles/liter, solubility of Ba^{2+} is s moles/liter and solubility of F^- is 2s moles/liter

Therefore,  

K_sp=[Ba^{2+}][F^{-}]^2

2.45\times 10^{-5}=[s][2s]^2

4s^3=2.45\times 10^{-5}

s^3=6.12\times 10^{-6}moles/liter

s=0.0183moles/liter

Thus the molar solubility of barium fluoride is 0.0183 moles/liter.

8 0
4 years ago
Three Stoichiometry Questions
andrezito [222]

Answer:

Explanation:

7)

Given data:

Mass of aluminium = 2.5 g

Mass of oxygen = 2.5 g

Mass of aluminium oxide = 3.5 g

Percent yield = ?

Solution:

Chemical equation:

4Al + 3O₂   →   2Al₂O₃

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 27 g/mol

Number of moles = 0.09 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 32 g/mol

Number of moles = 0.08 mol

Now we will compare the moles of aluminium oxide with aluminium and oxygen.

                          Al         ;       Al₂O₃

                           4         :        2

                        0.09      :       2/4×0.09 = 0.045

                          O₂       :        Al₂O₃

                          3         :          2

                         0.08    :        2/3 ×0.08 = 0.053

The number of moles of aluminium oxide produced by Al are less so it will limiting reactant.

Mass of aluminium oxide:

Mass = number of moles × molar mass

Mass = 0.045  × 101.96 g/mol

Mass = 4.6 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 3.5 g / 4.6 ×100

Percent yield = 76.1%

8)

Given data:

Mass of copper produced = 3.47 g

Mass of aluminium = 1.87 g

Percent yield = ?

Solution:

Chemical equation:

2Al + 3CuSO₄   →   Al₂(SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 1.87 g/ 27 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of copper with aluminium.

                          Al         ;       Cu

                           2         :        3

                        0.07      :       3/2×0.09 = 0.105

             

Mass of copper:

Mass = number of moles × molar mass

Mass = 0.105  × 63.55 g/mol

Mass = 6.67 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield =  3.47 g / 6.67 × 100

Percent yield = 52%

                       

4 0
3 years ago
student has calibrated his/her calorimeter and finds the heat capacity to be 14.2 J/°C. S/he then determines the molar heat capa
nika2105 [10]

Answer:

24.03 J/mol.ºC

Explanation:

For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.

-Qal = Qw + Qc (minus signal represents that the heat is lost)

-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc

Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC

-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)

1971.15Cal = 1699.10 + 58.22

1971.15Cal = 1757.32

Cal = 0.89 J/g.ºC

The molar mass of Al is 27 g/mol

Cal = 0.89 J/g.ºC * 27 g/mol

Cal = 24.03 J/mol.ºC

6 0
3 years ago
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