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2 years ago

How many moles of carbon dioxide gas should be produced when 10.0 g of C2H6 are combusted at STP?

Chemistry

1 answer:

Finger [1]2 years ago

3 0

Answer:

0.665 moles of CO₂

Explanation:

The balance chemical equation for the combustion of Ethane is as follow:

2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

Step 1: Calculate moles of C₂H₆ as;

Moles = Mass / M.Mass

Putting values,

Moles = 10.0 g / 30.07 g/mol

Moles = 0.3325 moles

Step 2: Calculate Moles of CO₂ as;

According to balance chemical equation,

2 moles of C₂H₆ produced = 4 moles of CO₂

So,

0.3325 moles of C₂H₆ will produce = X moles of CO₂

Solving for X,

X = 0.3325 mol × 4 mol ÷ 2 mol

X = 0.665 moles of CO₂

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

2 years ago

How many grams of mercury can be produced if 18.0 g of mercury (11) oxide decomposes?

NARA [144]

Answer:

18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

Explanation:

Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.

The decomposition of mercury oxide is given by the chemical equation below:

2HgO ----> 2Hg + O₂

2 moles of HgO decomposes to produce 1 mole of Hg

2 moles of HgO has a mass of 433.2 g

433.2 g of HgO produces 216.6 g of Hg

18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg

Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

3 0

2 years ago

What is the mass, in grams, of a liquid having a density of 1.50g/mL and a volume of 3500mL

mojhsa [17]

Answer:

5,250g

Explanation:

Density = Mass / Volume

*Note: mass = x

1.50g/mL = x / 3500mL

multiply 3500mL on both sides

1.50g/mL * 3500mL = 3500mL(x) / 3500mL

cancel units

x = 5250g

5 0

2 years ago

Which of the following an ionic bond like what holds Na+ and Cl- together in NaCl?

galina1969 [7]

Answer:

Explanation:

Rb look closely

8 0

3 years ago

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bagirrra123 [75]

Answer:

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8 0

2 years ago