The answer is option C.
That is it is a heterogeneous mixture.
Heterogeneous mixture have the following properties:
1. Different components could be observed in the substance.
2. Different samples of the substance appeared to have different proportions of the components.
3.The components could be easily separated using filters and sorting.
Answer:
5.5 atm
Explanation:
Step 1: Calculate the moles in 2.0 L of oxygen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
2.0 L × 1 mol/22.4 L = 0.089 mol
Step 2: Calculate the moles in 8.0 L of nitrogen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
8.0 L × 1 mol/22.4 L = 0.36 mol
Step 3: Calculate the total number of moles of the mixture
n = 0.089 mol + 0.36 mol = 0.45 mol
Step 4: Calculate the pressure exerted by the mixture
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 0.45 mol × (0.0821 atm.L/mol.K) × 298 K / 2.0 L = 5.5 atm
In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
<h3>What is Enthalpy of Vaporization ?</h3>
The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.
<h3>How to find the energy change from enthalpy of vaporization ?</h3>
To calculate the energy use this expression:
where,
Q = Energy change
n = number of moles
= Molar enthalpy of vaporization
Now find the number of moles
Number of moles (n) = 
= 
= 0.5 mol
Now put the values in above formula we get
[Negative sign is used because Br₂ condensed here]
= - (0.5 mol × 15.4 kJ/mol)
= - 7.7 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
Learn more about the Enthalpy of Vaporization here: brainly.com/question/13776849
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