Answer:
Explanation:
Ratio of mass of C , N , H and O
= .8007 :0.9333:0.2016:2.133
Ratio of moles of C , N , H and O
= .8007/12 : .9333 / 14 : 0.2016 / 1 : 2.133/16
= .0667 : .0667: .2016 : .1333
= .0667 / .0667 : .0667 / .0667 : .2016 /.0667 : .1333 / .0667
= 1 : 1 : 3: 2
Hence empirical formula = CNH₃O₂
7 .
Weight of titanium Ti = 1.916 g
Weight of oxygen = 3.196 - 1.916 = 1.28 g
Ratio of weight of Ti and O
= 1.916 : 1.28
Ratio of moles of Ti and O
1.916/48 : 1.28/16 [ Molecular weight of Titanium is 48 ]
= .04 : .08
= .04/.04 : .08/.04
= 1 :2 .
Empirical formula
TiO₂
What's the relationship between total and partial pressure? The total pressure is the sum of the parcial pressures!
So for us, it would be:
378= 212+101+x
where x is the parcial pressure of nitrogen.
Now we count:
378= 212+101+x
378=313+x
378-313=x
65=x
So the parcial pressure exerted by nitrogen is 65!
The heat energy that causes most of the water evaporate on the surface of the Earth is the Sun. Which provides earth most of the thermal energy into the hydro-thermodynamic system on Earth.
{{Sun}}
Answer:
0.15g
Explanation:
Given parameters:
Number of molecules of water = 1.2 x 10²¹ molecules
Unknown:
Mass of SnO₂ = ?
Solution:
To solve this problem, we have to work from the known to the unknown specie;
SnO₂ + 2H₂ → Sn + 2H₂O
Ensure that the equation given is balanced;
Now,
the known species is water;
6.02 x 10²³ molecules of water = 1 mole
1.2 x 10²¹ molecules of water = = 0.2 x 10⁻²moles
Number of moles of water = 0.002moles
From the balanced chemical equation:
2 mole of water is produced from 1 mole of SnO₂
0.002 moles of water will be produced from = 0.001moles
To find the mass;
Mass = number of moles x molar mass
Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol
Mass = 0.001 x 150.7 = 0.15g
Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ