Answer:
50 g of S are needed
Explanation:
To star this, we begin from the reaction:
S(s) + O₂ (g) → SO₂ (g)
If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.
According to stoichiometry, we can determine the moles of sulfur dioxide produced.
100 g. 1mol / 64.06g = 1.56 moles
This 1.56 moles were orginated by the same amount of S, according to stoichiometry.
Let's convert the moles to mass
1.56 mol . 32.06g / mol = 50 g
I think it’s 1 mole
If Iam not mistaken
Answer:
a) Pabs = 48960 KPa
b) T = 433.332 °C
Explanation:
∴ d = 1000 Kg/m³
∴ g = 9.8 m/s²
∴ h = 5000 m
∴ P gauge = - 40 KPa * ( 1000 Pa / KPa ) = - 40000 Pa; Pa≡Kg/m*s²
⇒ Pabs = - 40000 Kg/ms² + ( 1000 Kg/m³ * 9.8 m/s² * 5000 m )
⇒ Pabs = 48960000 Pa = 48960 KPa
a) at that height and pressure, we find the temperature at which the water boils by means of an almost-exponential graph which has the following equation:
P(T) = 0.61094 exp ( 17.625*T / ( T + 243.04 ))......P (KPa) ∧ T (°C)....from literature
∴ P = 48960 KPa
⇒ ( 48960 KPa / 0.61094 ) = exp ( 17.625T / (T+ 243.04))
⇒ 80138.803 = exp ( 17.625T / ( T + 243.04))
⇒ Ln ( 80138.803) = 17.625T / ( T + 243.04))
⇒ 11.292 * ( T + 243.04 ) = 17.625T
⇒ 11.292T + 2744.289 = 17.625T
⇒ 2744.289 = 17.625T - 11.292T
⇒ 2744.289 = 6.333T
⇒ T = 433.332 °C
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
