58.08 g/mol
Formula: C3H6O
Melting point: -139°F (-95°C)
Density: 784 kg/m³
Gold (III) nitrate in an aqueous solution is hydrolyzed with formation of gold (III) metahydroxide.
Au(NO₃)₃ → Au³⁺(aq) + 3NO₃⁻(aq)
Au³⁺ + H₂O ⇄ AuOH²⁺ + H⁺
AuOH²⁺ + H₂O ⇄ Au(OH)₂⁺ + H⁺
Au(OH)₂⁺ + H₂O → AuOOH·H₂O(s) + H⁺
Au(NO₃)₃(aq) + 2H₂O(l) = AuOOH(s) + 3HNO₃(aq)
Answer:
The molecular formule for this unknow molecule is C2H4O2
Explanation:
The empirical formula is CH2O ( or better said CnH2nOn)
This means there are 3 elements in the formula of this molecule
⇒ Carbon (C) with a Molar mass of 12 g/mole
⇒ Hydrogen (H) with a Molar mass of 1 g/mole
⇒ Oxygen (O) with a Molar mass of 16 g/mole
We can also notice that the amount of hydrogen should 2x the amount of carbon ( also 2x the amount of oxygen).
The mass of the empirical formule = 12g/ mole + 2* 1 g/mole + 16 g/mole = 30 g/mole
To know what number is n in CnH2nOn we should divide the molecular mass by the empirical mass:
60 g/mole / 30g/mole = 2
this means n = 2
and this will give a molecular formule of C2H4O2
We can control this to calculate the molecular mass:
2*12 + 4* 1 + 2*16 = 24 + 4 + 32 = 60 g/mole
The molecular formule for this unknow molecule is C2H4O2
Answer:
0.06mole
Explanation:
Given parameters:
Volume of balloon = 2.5L
T = 343.5K
P = 66.7kPa; in atm gives = 0.66atm
Unknown:
Number of moles of He = ?
Solution:
To solve this problem, we use the ideal gas equation.
It is mathematically expressed as;
Pv = nRT
where p is the pressure
v is the volume
n is the number of moles
R is the gas constant
T is the temperature
0.66 x 2.5 = n x 0.082 x 343.5
n = 0.06mole
Answer:
The yield is 20 %.
Explanation:
Mg + 2 HCl → MgCl2 + H2
(12 g Mg) / (24.30506 g Mg/mol) x (1 mol H2 / 1 mol Mg) x (2.015894 g H2/mol) = 0.9953 g H2.
(0.2 g) / (0.9953 g) = 0.20 = 20% yield
