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monitta
4 years ago
15

Nitrogen gas (N₂) reacts with hydrogen gas (H₂) to form ammonia (NH₃). At 200°C in a closed container, 1.0 atm of nitrogen gas i

s mixed with 2.0 atm of hydrogen gas. At equilibrium, the total pressure is 1.9 atm. Calculate the partial pressure of hydrogen gas at equilibrium.
Chemistry
1 answer:
Pie4 years ago
8 0

Answer:

The partial pressure of hydrogen gas at equilibrium is 1.26 atm

Explanation:

Let's use the molar fraction to solve this:

Molar fraction = Moles of gas / Total moles

Molar fraction = Gas pressure / Total pressure

Without equilibrium, we can think that the total system pressure is the sum of the partial pressures of each gas.

1 atm N₂ + 2 atm H₂ = 3 atm

Molar fraction for H₂ = 2 atm / 3atm → 0.66

Let's replace the molar fraction in equilibrium

Gas pressure / 1.9 atm = 0.66

Gas pressure = 1.26atm

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Carbon dioxide

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5 0
3 years ago
A 10 pound block of lead-210 has a half-life of 22 years. What percentage of the block still contains lead-210 after 60 years?
avanturin [10]

Answer:

The percentage of the block contains 15% after 60 years.

Explanation:

Step 1: Formula for half-life time

To calculate the half-life time we will use the following formula:

At = A0 * 1/2 ^(x/t)

With At = the quantity after a time t

A0 = The quantity at time t = 0 (start)

x = time in this case = 60 years

t= half-life time = 22 years

Step 2: Calculate the percentage after 60 years

In this case: after 60 years the percentage will be

A0= 10 * 1/2 ^(60/22)

A0 = 1.5

A0 / At  = 1.5 /10 = 0.15

0.15 *100% = 15 %

The percentage of the block contains 15% after 60 years.

7 0
3 years ago
The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
4 years ago
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