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.
Answer:
a. 50ml b.10ml c. 6.097ml d. 190.1 ml
Explanation:
According to Boyle's law
Volume is inversely proportional to pressure at constant temerature
Mathematically
P1V1=P2V2
P1=Initial pressure=0.8atm
V1=Initial volume=25ml
making V2 the subject
at 0.4atm P2=0.4 atm,
V2=25×0.8/0.4
=50ml
at 2 atm V2=25×0.8/2
=10 ml
1mmHg=0.00131579
2500mmHg=3.28 atm
At 3.28 atm,V2=25×0.8/3.28
=6.097 ml
at 80.0 torr
1 torr=0.00131579
80 torr=0.1052 atm
at 0.1048 atm V2=25×0.8/0.1048
=190.1 ml
Nucleus ,endoplasmic reticulum
