Cl-35 occurs in greater abundance.
The <em>weighted atomic mass</em> lies <em>closer to 35 u</em> than to 37 u, so the Cl-35 isotope must be more abundant.
Using the ideal gas law equation, we can find the number of H₂ moles produced.
PV = nRT
Where P - pressure - 0.811 atm x 101 325 Pa/atm = 82 175 Pa
V - volume - 58.0 x 10⁻³ m³
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 32 °C + 273 = 305 K
substituting these values in the equation,
82 175 Pa x 58.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 305 K
n = 1.88 mol
The balanced equation for the reaction is as follows;
CaH₂(s) + 2H₂O(l) --> Ca(OH)₂(aq) + 2H₂(g)
stoichiometry of CaH₂ to H₂ is 1:2
When 1.88 mol of H₂ is formed , number of CaH₂ moles reacted = 1.88/2 mol
therefore number of CaH₂ moles reacted = 0.94 mol
Mass of CaH₂ reacted - 0.94 mol x 42 g/mol = 39.48 g of CaH₂ are needed
Answer:
Explanation:
We can use the Combined Gas Laws to solve this problem
Data
p₁ = 571.2 Torr; p₂ = 400 Torr
V₁ = 3.5 L; V₂ = ?
T₁ = 21.5 °C; T₂ = 6.8 °C
Calculations
(a) Convert the temperatures to kelvins
T₁ = (21.55 + 273.15) K = 294.70 K
T₂ = (6.8 + 273.15) K = 279.95 K
(b) Calculate the new volume
Hope this helps solve your problem!
Answer:
2.52kg
Explanation:
80mm = 8 cm
0.15m = 15 cm
volume of the container = 10m × 8cm × 15cm
= 1200 cubic centimetres
mass = density × volume
= 2.1 × 1200
= 2520g
2520/ 1000
= 2.52kilograms
