Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
Answer:
for what I can see in the picture the volume is 155
Explanation:
To solve this problem,
we can use the Henderson-Hasselbalch Equation which relates the pH to the measure
of acidity pKa. The equation is given as:<span>
<span>pH = pKa + log ([base]/[acid]) ---> 1</span></span>
Where,
[base] = concentration
of C2H3O2
in molarity or moles
<span>[acid] = concentration of HC2H3O2 in molarity or moles</span>
For the sake of easy calculation, let us assume that:
[base] = 1
[acid] = x
<span>
Therefore using equation 1,
4.24 = 4.74 + log (1 / x)
<span>log (1 / x) = - 0.5
1 / x = 0.6065 </span></span>
x =
1.65<span>
The required ratio of C2H3O2 /HC2H3O2 <span>
is 1:1.65 or 3:5. </span></span>
Answer:
Explanation:
Hello,
In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:
Therefore the equilibrium temperature shows up as:
Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:
Best regards.
The heat absorbed by the water is
Q = 500 (4.18) (32.2 - 25)
Q = 15048 J
The enthalpy of fusion of the sodium acetate is:
<span>ΔHf = Q / m
</span><span>ΔHf = 15048 / 100
</span>ΔHf = 150.48 J/g
