Answer:
KO is the limiting reactant.
0.11 mol O₂ will be produced.
Explanation:
4 KO₂ + 2 H₂O ⇒ 4 KOH + 3 O₂
Find the limiting reagent by dividing the moles of the reactant by the coefficient in the equation.
(0.15 mol KO₂)/4 = 0.0375
(0.10 mol H₂O)/2 = 0.05
KO₂ is the limiting reagent.
The amount of product produced depends on the limiting reagent. To find how much is produced, take moles of limiting reagent and multiply it by the ratio of reagent to product. You can find the ratio by looking at the equation. For every 4 moles of KO₂, 3 moles of O₂ are produced.
0.15 mol KO₂ (3 mol O₂)/(4 mol KO₂) = 0.1125 mol O₂
0.11 mol O are produced.
Answer:
Element; compound
Explanation:
Oxygen is an element on the periodic table, but when it's combined with the element Hydrogen, it produces water, which is a compound (H2O). When two elements combine, it produces a compound.
I hope this helps! :)
Answer:
One gram helium gas is equal to the 4.002602 g/mole
Explanation:
Helium gas is a gas or chemical element. The symbol of helium gas is He and the atomic number of this gas is 2. This gas is colorless, odorless, and tasteless. It is a non-toxic gas. It is found in the periodic table in the first place.
Helium is a very lightest gas and very abundant in nature in the universe. It is present in the universe around 24 percent. Helium is also called by another name that titan of sun. This is detected by the yellow line. The liquid form of helium gas is called cryogenics. This helium gas is used in lifting balloons and airplanes.
Answer:
Molar mass for the unknown solute is 109 g/mol
Explanation:
Freezing point depression is the colligative property that must be applied to solve the question.
T°F pure solvent - T°F solution = Kf . m
Let's analyse the data given
Camphor → solvent
Unknown solute → The mass we used is 0.186g
T°F pure solvent = 179.8°C and T°F solution = 176.7°C.
These data help us to determine the ΔT → 179.8°C - 176.7°C = 3.1°C
So we can replace → 3.1°C = 40°C/m . m
m = 3.1°C / 40 m/°C → 0.0775 mol/kg
We have these moles of solute in 1kg of solvent, but our mass of camphor is 22.01 g (0.02201 kg).
We can determine the moles of solute → molality . kg
0.0775 mol/kg . 0.02201 kg = 1.70×10⁻³ moles
Molar mass → mass (g) / moles → 0.186 g / 1.70×10⁻³ mol = 109 g/mol
