Answer:
Explanation:
Oxygen is one of the most abundant elements on this planet. Our atmosphere is 21% free elemental oxygen. Oxygen is also extensively combined in compounds in the earths crust, such as water (89%) and in mineral oxides. Even the human body is 65% oxygen by mass.
Free elemental oxygen occurs naturally as a gas in the form of diatomic molecules, O2 (g). Oxygen exhibits many unique physical and chemical properties. For example, oxygen is a colorless and odorless gas, with a density greater than that of air, and a very low solubility in water. In fact, the latter two properties greatly facilitate the collection of oxygen in this lab. Among the unique chemical properties of oxygen are its ability to support respiration in plants and animals, and its ability to support combustion.
In this lab, oxygen will be generated as a product of the decomposition of hydrogen peroxide. A catalyst is used to speed up the rate of the decomposition reaction, which would otherwise be too slow to use as a source of oxygen. The catalyst does not get consumed by the reaction, and can be collected for re-use once the reaction is complete. The particular catalyst used in this lab is manganese(IV) oxide.
Answer:
H2O
Explanation:
There are 2 hydrogen atoms and 1 oxygen atom
Solution of 0.25 M is prepared in two steps,
1) Calculate Amount of Solute:
Molar Mass of Solute: 342.3 g/mol
As we know,
Molarity = Moles / 1 dm³
or,
Moles = Molarity × 1 dm³
Putting Values,
Moles = 0.25 mol.dm⁻³ × 1 dm³
Moles = 0.25 moles
Now, find out mass of sucrose,
As,
Moles = Mass / M.mass
or,
Mass = Moles × M.mass
Putting Values,
Mass = 0.25 mol × 342.3 g.mol⁻¹
Mass = 85.57 g
2) Prepare Solution:
Take Volumetric flask and add 85.57 g of sucrose in it. Then add distilled water up to the mark of 1 dm³. Shake well! The solution prepared is 0.25 M in 1 Liter.
A solute is the substance to be dissolved for example sugar
The solvent is the one doing the dissolving for example water
Answer:
Final pressure is 1.42atm
Explanation:
Based on Gay-Lussac's law, pressure of a gas is directely proportional to its absolute temperature. The equation of this law is:
P₁T₂ = P₂T₁
<em>Where P is pressure and T is absolute temperature of 1, initial state and 2, final state of the gas.</em>
In the problem, initial conditions are Standard Temperature and Pressure, STP, that are 1 atm and 273.15K.
If the final temperature is 115°C = 388.15K (115°C + 273.15 = 388.15K), using Gay-Lussac's law:
P₁T₂ = P₂T₁
1atmₓ388.15K = P₂ₓ273.15K
1.42atm = P₂
<h3>Final pressure is 1.42atm</h3>
<em />
<em />