Answer:
172 g Al
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 26.98 101.96
4Al + 3O₂ ⟶ 2Al₂O₃
m/g: 325
(a) Calculate the <em>moles of Al₂O₃
</em>
n = 325 g Al₂O₃ × 1 mol Al₂O₃ /39.10 g Al₂O₃
n = 3.188 mol Al₂O₃
(b) Calculate the <em>moles of Al
</em>
The molar ratio is (4 mol Al/2 mol Al₂O₃)
n = 3.188 mol Al₂O₃ × (4 mol Al/2 mol Al₂O₃)
n = 6.375 mol Al
(c) Calculate the <em>mass of Al</em>
m = 6.375 mol Al × (26.98 g Al/1 mol Al)
m = 172 g Al
Note: The answer can have only <em>three</em> significant figures because that is all you gave for the mass of Al₂O₃.
The lower you go, the more acidic. The higher you go, the more alkaline. Your answer would most likely be 6.
Answer: 5.66 dm3
Explanation:
Given that:
Volume of neon gas = ?
Temperature T = 35°C
Convert Celsius to Kelvin
(35°C + 273 = 308K)
Pressure P = 0.37 atm
Number of moles N = 0.83 moles
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
0.37atm x V = 0.83 moles x 0.0082 atm dm3 K-1 mol-1 x 308K
0.37 atm x V = 2.096 atm dm3
V = (2.096 atm dm3 / 0.37atm)
V = 5.66 dm3
Thus, the volume of the neon gas is 5.66 dm3
Answer:
Saturated solution = 180 gram
Explanation:
Given:
Solubility of Z = 60 g / 100 g water
Given temperature = 20°C
Amount of water = 300 grams
Find:
Saturated solution
Computation:
Saturated solution = [Solubility of Z] × Amount of water
Saturated solution = [60 g / 100 g] × 300 grams
Saturated solution = [0.6] × 300 grams
Saturated solution = 180 gram
Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)
Answer:
-3670.33 J/K
Explanation:
Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.
Mathematically, change of Entropy can be expressed as,
ΔS = ΔH/T ....................................... Equation 1
Where ΔS = Change of entropy, ΔH = heat change, T = temperature.
ΔH = -(Lf×m).................................... Equation 2
Note: ΔH is negative because heat is lost.
Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg
Substitute into equation
ΔH = -(3.34×10⁵×3.0)
ΔH = - 1002000 J.
But T = 0 °C = (0+273) K = 273 K.
Substitute into equation 1
ΔS = -1002000/273
ΔS = -3670.33 J/K
Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C
