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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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Which of the following is the method that incorporates designing sustainable living environments with agriculture?
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3 years ago
Given the half‑reactions and their respective standard reduction potentials 1. Cr 3 + + e − ⟶ Cr 2 + E ∘ 1 = − 0.407 V 2. Cr 2 +
Licemer1 [7]

Answer:

The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is -0.744 V

Explanation:

Here we have

1. Cr³⁺  + e − ⟶ Cr²⁺ E⁰₁ = − 0.407 V

2. Cr²⁺ + 2 e − ⟶ Cr ( s ) E⁰₂  = − 0.913 V

To solve the question, we convert, the E⁰ values to ΔG as follows

ΔG₁ = n·F·E⁰₁ and ΔG₂ = n·F·E⁰₂

Where:

F = Faraday's constant in calories

n = Number of e⁻

ΔG₁ = Gibbs free energy for the first reaction

ΔG₂ = Gibbs free energy for the second half reaction

E⁰₁  = Reduction potential for the first half reaction

E⁰₂ = Reduction potential for the second half reaction

∴ ΔG₁ = 1 × F × − 0.407 V

ΔG₂ = 2 × F  × − 0.913 V

ΔG₁  + ΔG₂  = F × -2.233 V which gives

ΔG = n × F × ΔE⁰ = F × -2.233 V  

Where n = total number of electrons ⇒ 1·e⁻ + 2·e⁻ = 3·e⁻ = 3 electrons

We have, 3 × F × ΔE⁰ = F × -2.233 V

Which gives ΔE⁰ = -2.233 V /3 = -0.744 V.

7 0
3 years ago
Answer these please ASAP need help no idea how to do these
STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass =  g

Cl₂:

Number of moles = Mass / molar masa

Number of moles  = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol  = Mass / 2 g/mol

Mass =  16 g

P₄:

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles  = 1.6 g /48  g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles  = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles  = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles  = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO    →    2Fe  + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

                        Fe          :           F₂O₃                

                           2          :             1

                          1.78       :        1/2×1.78 = 0.89 mol

Mass of  F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂   →  2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

                          Al            :         AlI₃    

                          2             :           2

                         0.05         :        0.05

                           I₂            :         AlI₃

                           3            :          2

                         0.1           :           2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:                            

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂   → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

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

               Pb         :        O₂

                2          :         1

               0.03     :      1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass =  0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂   → 2PbO

   

6 0
3 years ago
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