Explanation:
from the equation 1 mole of O2 will give 2 moles of H2O then 6.0 moles of O2 will give x
6.0*2 moles/ 1 mole
= 12 moles
this implies that, 6.0 moles of O2 will give = 12 moles of water
Answer:B, the colloid particles are larger.
PLEASE GIVE BRAINLIEST AND WATCH OUT BEHIND YOU
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
Think of w shape with 4 sides the shape I know tat is Quadrilateral is a square
Answer:
127°C
Explanation:
This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.
We convert our value to K → -73°C + 273 = 200 K
The moles are the same, and the volume is also the same:
P₁ / T₁ = P₂ / T₂
But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂
P₁ / 200K = 2P₁ / T₂
1 /2OOK = (2P₁ / T₂) / P₁
See how's P₁ term is cancelled.
200K⁻¹ = 2/ T₂
T₂ = 2 / 200K⁻¹ → 400K
We convert the T° to C → 400 K - 273 = 127°C
