Answer:
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Answer: 1.10x10²³ atoms of C
110202600000000000000000 atoms C
Explanation:The solution process is shown below.
0.183 mole C x 6.022x10²³ atoms C / 1 mole C
= 1.10x10²³ atoms C
or 110202600000000000000000 atoms C
Colligative properties calculations are used for this type of problem. Calculations are as follows:
ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
<span>ΔT(boiling point) = (Kb)m
</span>m = 1.02 °C / 0.512 °C kg / mol
<span>m = 1.99 mol / kg
</span><span>ΔT(freezing point) = (Kf)m
</span>ΔT(freezing point) = 1.86 °C kg / mol (<span>1.99 mol / kg)
</span>ΔT(freezing point) = 3.70 <span>°C
</span>Tf - T = 3.70 <span>°C
T = -3.70 </span><span>°C</span>
Answer:
The magnesium reacted with the oxygen in the air.
Explanation:
For argument’s sake, let’s say that the mass of magnesium oxide was 3 g and that of the oxide was 5 g.
The reaction was
magnesium + oxygen ⟶ magnesium oxide
Mass: 3 g 5 g
Mass of oxygen = 5 g – 3 g = 2 g
The 3 g of magnesium must have combined with 2 g of oxygen to form 5 g of magnesium oxide.
<u>Answer:</u> The standard Gibbs free energy of the given reaction is 6.84 kJ
<u>Explanation:</u>
For the given chemical equation:
The expression of 
for above equation follows:
We are given:
Putting values in above expression, we get:
To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:
where,
= standard Gibbs free energy = ?
R = Gas constant = 8.314 J/K mol
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
= equilibrium constant at 25°C = 0.0632
Putting values in above equation, we get:
Hence, the standard Gibbs free energy of the given reaction is 6.84 kJ
