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pychu [463]
1 year ago
10

What volume of oxygen at STP is requieres for the complete combustion of 100.50 mL of C2H2

Chemistry
1 answer:
malfutka [58]1 year ago
7 0

The volume of oxygen at STP required would be 252.0 mL.

<h3>Stoichiometic problem</h3>

The equation for the complete combustion of C2H2 is as below:

2C_2H_2 + 5O_2 --- > 4CO_2 + 2H_2O

The mole ratio of C2H2 to O2 is 2:5.

1 mole of a gas at STP is 22.4 L.

At STP, 100.50 mL of C2H2 will be:

                 100.50 x 1/22400 = 0.0045 mole

Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles

0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL

Thus, 252.0 mL of O2 gas will be required at STP.

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

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Explanation:

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<em>Using cross multiplication:</em>

4 moles of NaOH produced with → 1 mole of O₂ .

15.7 moles of NaOH produced with → ??? mole of O₂ .

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8 0
3 years ago
if a sample of gas at 25.2 c has a volume of 536mL at 637 torr, what will its volume be if the pressure is increased to 712 torr
nignag [31]
Considering ideal gas:
PV= RTn

T= 25.2°C = 298.2 K

P1= 637 torr = 0.8382 atm

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:. R=0.082 atm.L/K.mol

:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
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:. n= O.0184 mol

Then,
P2= 712 torr = 0.936842 atm

V2 = RTn/P2 = [(0.082atmL/
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6 0
3 years ago
The decay of a living things allows chemical elements to be lost.
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Answer: is is true

Explanation:

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There have been different models of the atom over time. How has the competition between these models affected our understanding
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Doing the same final thanks 
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An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 300 nm and then immediately emits photon whose ass
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Answer:

The net energy is 2.196 eV

Explanation:

Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.

Using:

ΔE = h*c*(1/λ_{1} - 1/λ_{2})

where:

ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*10^{-19} J.

h = 4.135*10^{-15} eVs

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Thus:

ΔE = 4.135*10^{-15} eVs*3*10^{8} m/s*(\frac{1}{300*10^{-9}m } }-\frac{1}{640*10^{-9}m })

ΔE = 4.135*10^{-15}*3*10^{8}*1.77*10^{6} eV = 2.196 eV

6 0
3 years ago
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