Answer:
d = 0.98 g/L
Explanation:
Given data:
Density of acetylene = ?
Pressure = 0.910 atm
Temperature = 20°C (20+273 = 293 K)
Solution:
Formula:
PM = dRT
R = general gas constant = 0.0821 atm.L/mol.K
M = molecular mass = 26.04 g/mol
0.910 atm × 26.04 g/mol = d × 0.0821 atm.L/mol.K×293 K
23.7 atm.g/mol = d × 24.1 atm.L/mol
d = 23.7 atm.g/mol / 24.1 atm.L/mol
d = 0.98 g/L
Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Answer:
The mass of oxygen is 12.10 g.
Explanation:
The decomposition reaction of potassium chlorate is the following:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
We need to find the number of moles of KClO₃:
Where:
m: is the mass = 30.86 g
M: is the molar mass = 122.55 g/mol
Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3
Finally, the mass of O₂ is:
Therefore, the mass of oxygen is 12.10 g.
I hope it helps you!
That is correct c
Explanation
Answer:
no examination in 16.9g in molicube i n gas
Explanation:
sana po makatulong po sa inyo
