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nignag [31]
3 years ago
14

What volume. In liters, of H2O(g) measured at STP is produced by the combustion of 15.63 g of natural gas (CH4) according to the

following equation? CHale) +20269) CO2 + 2H2008)
Chemistry
1 answer:
fomenos3 years ago
3 0

Answer:

V = 43.95 L

Explanation:

Given data:

Mass of CH₄ decomposed = 15.63 g

Volume of H₂O produced at STP = ?

Solution:

Chemical equation:

CH₄ + 2O₂    →       2H₂O  + CO₂

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 15.63 g/ 16 g/mol

Number of moles = 0.98 mol

Now we will compare the moles of H₂O with CH₄.

                         CH₄              :              H₂O

                           1                 :                2

                        0.98             :            2×0.98 = 1.96 mol

Volume of hydrogen:

PV = nRT

1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K

V = 43.95atm.L / 1atm

V = 43.95 L

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What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the
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Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

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Equation of the process;

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Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

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Ecell= E°cell= 1.05 V

3 0
3 years ago
How many moles are in 1.2×10^3g of ammonia
Diano4ka-milaya [45]

Answer:8

Explanation:

4 0
3 years ago
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In the following reaction, how many liters of oxygen produce 560 liters of
saw5 [17]

The balanced chemical reaction will be:

CH4 + 2O2 → CO2 + 2H2O

We are given the amount of carbon dioxide to produce from the reaction. This will be our starting point.

 

560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) ( 22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>

7 0
3 years ago
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Molecules have ionization energies, just like atoms do. The ionization energy for molecular oxygen, for example, is 1314 kJ/mol.
taurus [48]

Answer:

91.1835 nm

Explanation:

Given that the ionization energy of the oxygen molecule = 1314 kJ/mol

It means that

1 mole of oxygen molecules can be ionized by the energy = 1314 kJ = 1314000 J

1 mole of molecules contains 6.022 × 10²³ atoms

So,

6.022 × 10²³ atoms of oxygen molecules can be ionized by the energy = 1314000 J

1 atom require \frac{1314000}{6.022\times 10^{23}}\ J of energy

Energy = 2.18\times 10^{-18}\ J

Also

E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength

So,

\lambda=\frac {h\times c}{E}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{2.18\times 10^{-18}}\ m

\lambda=91.1835\times 10^{-9}}\ m

Also,

1\ nm=10^{-9}}\ m

So, wavelength = 91.1835 nm

6 0
3 years ago
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