Answer:
194.6 mL of SO₂
Explanation:
The reaction that takes place is:
P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)
<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:
- 23.8 °C → 23.8 + 273.15 = 296.95 K
- 747 torr → 747/760 = 0.983 atm
We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):
0.576 g P₄S₃ * 
= 7.85 * 10⁻³ mol SO₂ = n
PV=nRT
0.983 atm * V = 7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K
V = 0.1946 L
- Finally we convert L into mL:
0.1946 * 1000 = 194.6 mL
Answer:
Chopping wood logs
A pot of water o a grate over a burning fire
Explanation:
Physical change is easily reversible. Burning/combustion is a chemical process where substances react rapidly with oxygen: this is usually irreversible.
The marshmallow, roasted food and burned wood all undergo combustion and hence are tagged chemical changes.
Explanation:
Answer:
The artifact is 570 years old. That is, 5.7 × 10² years.
Explanation:
Radioactive decay follows first order reaction kinetics.
Let the initial activity for fresh Carbon-14 be A₀
And the activity at any other time be A
The rate of radioactive decay is given by
dA/dt = - KA
dA/A = - kdt
Integrating the left hand side from A₀ to A₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)
In [(A₀/2)/A₀] = - k t(1/2)
In (1/2) = - k t(1/2)
- In 2 = - k t(1/2)
k = (In 2)/t₍₁,₂₎
t(1/2) is given in the question to be 5.73 × 10³ years
k = (In 2)/5730 = 0.000121 /year
dA/A = - kdt
Integrating the left hand side from A₀ to A and the right hand side from 0 to t
In (A/A₀) = - kt
A/A₀ = e⁻ᵏᵗ
A = A₀ e⁻ᵏᵗ
A = 2.8 × 10³ Bq.
A₀ = 3.0 × 10³ Bq.
2.8 × 10³ = 3.0 × 10³ e⁻ᵏᵗ
0.9333 = e⁻ᵏᵗ
e⁻ᵏᵗ = 0.9333
-kt = In 0.9333
- kt = - 0.06899
t = 0.06899/0.000121 = 570.2 years = 5.7 × 10² years
Answer:
5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.
Explanation:
The balanced reaction:
CO (g) + 2 H₂ (g) -> CH₃OH (l)
By stoichiometry of the reaction, the following amounts of moles of each compound participate in the reaction:
- CO: 1 mole
- H₂: 2 moles
- CH₃OH: 1 mole
Being the molar mass of each compound:
- CO: 28 g/mole
- H₂: 1 g/mole
- CH₃OH: 32 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CO: 1 mole* 28 g/mole= 28 grams
- H₂: 2 moles* 1 g/mole= 2 grams
- CH₃OH: 1 mole* 32 g/mole= 32 grams
Being 6 kg equivalent to 6000 grams (1 kg= 1000 grams), you can apply the following rules of three:
- If by stoichiometry 32 grams of methanol are formed from 28 grams of carbon monoxide, 6000 grams of methanol are formed from how much mass of carbon monoxide?
mass of carbon monoxide= 5250 grams= 5.25 kg
If by stoichiometry 32 grams of methanol are formed from 2 grams of hydrogen, 6000 grams of methanol are formed from how much mass of hydrogen?
mass of hydrogen= 375 grams
<u><em>5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol. </em></u>
