Answer:
Final concentrations:
Cu²⁺ = 0
Al³⁺ = 3.13 mmol/L = 84.51 mg/L
Cu = 4.7 mmol/L = 300 mg/L
Al = 0.57 mmol/L = 15.49 mg/L
Explanation:
2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)
Al: 27 g/mol ∴ 100 mg = 3.7 mmol
Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol
3 mol Cu²⁺ _______ 2 mol Al
4.7 mmol Cu²⁺ _____ x
x = 3.13 mmol Al
4.7 mmol of Cu²⁺ will be consumed.
3.13 mmol of Al will be consumed.
4.7 mmol of Cu will be produced.
3.13 mmol of Al³⁺ will be produced.
0.57 mmol of Al will remain.
Yes, mass never changes. No exceptions.
Answer:
It is rich in organisms because sunlight passes through its shallow water enabling photosynthesis to occur.
Explanation:
I hope that helped!!
Answer:
65.2L
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in this question,
P = 1.631 atm
V = ?
n = 4.3 moles
T = 28°C = 28 + 273 = 301K
Using PV = nRT
V = nRT/P
V = 4.3 × 0.0821 × 301 ÷ 1.631
V = 106.26 ÷ 1.631
V = 65.15
Volume of the gas = 65.2L
We need to keep in mind that the compound is neutral.
H2SO3
2(+1)+S+3(-2)=0 (since its neutral)
2+S-6=0
S-4=0
S=4
Therefore the oxidation number for sulfur is +4.
