Answer:
[N₂] = 0.032 M
[O₂] = 0.0086 M
Explanation:
Ideal Gas Law → P . V = n . R . T
We assume that the mixture of air occupies a volume of 1 L
78% N₂ → Mole fraction of N₂ = 0.78
21% O₂ → Mole fraction of O₂ = 0.21
1% another gases → Mole fraction of another gases = 0.01
In a mixture, the total pressure of the system refers to total moles of the mixture
1 atm . 1L = n . 0.082L.atm/mol.K . 298K
n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles
We apply the mole fraction to determine the moles
N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂
O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂
It makes your clothes smell great
Answer:
142.82 g
Explanation:
The following data were obtained from the question:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Density of gol= 19.3 g/cm³
Mass of gold =.?
Next, we shall determine the volume of the gold. This can be obtained as follow:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Volume of gold =.?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 19.4 – 12
Volume of gold = 7.4 mL
Finally, we shall determine the mass of the gold as follow:
Note: 1 mL is equivalent to 1 cm³
Volume of gold = 7.4 mL
Density of gol= 19.3 g/cm³ = 19.3 g/mL
Mass of gold =?
Density = mass /volume
19.3 = mass of gold /7.4
Cross multiply
Mass of gold = 19.3 × 7.4
Mass of gold = 142.82 g
Therefore, the mass of the gold pebble is 142.82 g
The answer Is B;A chemical bond holds atoms together.
We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.
mass of Xenon (Xe) = 5.08 g
mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g
Determine the number of moles of each of the element in the compound.
moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F
The empirical formula is therefore,
Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
<em> XeF₆</em>
