Answer:
Explanation:
Given that,
Area of sheet of Aluminium foil is 1 m²
Mass of the sheet = 3.636 g
The density of Aluminium, 
We need to find the thickness of the sheet in millimeters.
The density of an object is given in terms of its mass and volume as follows :
V = volume, V = A×t, t = thickness of the sheet
So,
Since, 1 cm = 10 mm
So,
t = 0.00134 mm
or
Some physical property's that describe how something feels is texture.
Answer:
E° = 0.00 V
E = 0.079 V
Explanation:
We can identify both half-reactions occurring in a concentration cell.
Anode (oxidation): Al(s) → Al³⁺(1.0 × 10⁻⁵ M) + 3 e⁻ E°red = -1.66 V
Cathode (reduction): Al³⁺(0.100 M) + 3 e⁻ → Al(s) E°red = -1.66 V
The global reaction is:
Al(s) + Al³⁺(0.100 M) → Al³⁺(1.0 × 10⁻⁵ M) + Al(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = -1.66 V - (-1.66 V) = 0.00 V
To calculate the cell potential (E) we have to use the Nernst equation.
E = E° - (0.05916/n) .log Q
where,
n: moles of electrons transferred
Q: reaction quotient
E = 0.00 V - (0.05916/3) .log (1.0 × 10⁻⁵/0.100)
E = 0.079 V
Answer:
680 g/m is the molar mass for the unknown, non electrolyte, compound.
Explanation:
Let's apply the formula for osmotic pressure
π = Molarity . R . T
T = T° absolute (in K)
R = Universal constant gases
π = Pressure
Molarity = mol/L
As units of R are L.atm/mol.K, we have to convert the mmHg to atm
760 mmHg is 1 atm
28.1 mmHg is (28.1 .1)/760 = 0.0369 atm
0.0369 atm = M . 0.082 L.atm/mol.K . 293K
(0.0369 atm / 0.082 mol.K/L.atm . 293K) = M
0.0015 mol/L = Molarity
This data means the mol of solute in 1L, but we have 100mL so
Molarity . volume = mol
0.0015 mol/L . 0.1L = 1.5x10⁻⁴ mole
The molar mass will be: 0.102g / 1.5x10⁻⁴ m = 680 g/m
Number 1 is 4.875 Number 2 is 3.47 repeating
