Pure- table salt
Impure- vegetable oil
The molarity of a solution that contains 20.0grams of sodium hydroxide in 200. ml of solution is 2.3 M.
<h3>What is Molarity? </h3>
Molarity is defined as the ratio of number of moles of solute to the volume of the solution.
Molarity = number of moles of solute/ Volume of solution
Given,
given mass of sodium hydroxide= 20 g
Volume of the solution = 200 mL.
Firstly, we will calculate the moles of sodium hydroxide.
<h3>What is Mole? </h3>
Mole is defined as the ratio of given mass of substance to the molar mass of substance.
Mole = given mass/ molar mass
Molar mass of sodium hydroxide = 40 g
Mole = 20/40
= 0.5 mol
Molarity = (0.5/ 200) × 1000 = 2.5M
Thus, we calculated that the molarity of a solution that contains 20.0grams of sodium hydroxide in 200. ml of solution is 2.3 M.
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The total kinetic energy of the gas sample is 3.3 KJ
<h3>What is kinetic energy? </h3>
This is the energy possessed by an object in motion. Mathematically, it can be expressed as:
KE = ½mv²
Where
- KE is the kinetic energy
- m is the mass
- v is the velocity
<h3>How to determine the mass of the fluorine gas</h3>
- Molar mass of fluorine gas = 38 g/mol
- Mole of fluorine gas = 1 mole
- Mass of fluorine gas = ?
Mass = mole × molar mass
Mass of fluorine gas = 1 × 38
Mass of fluorine gas = 38 g
<h3>How to determine the KE of the gas sample</h3>
- Mass (m) = 38 g = 38 / 1000 = 0.038 Kg
- Velocity (v) = 415 m/s
- Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.038 × 415²
KE = 3272.275 J
Divide by 1000 to express in kilojoule
KE = 3272.275 / 1000
KE = 3.3 KJ
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Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
<span>I believe that in this
case the controlled parameter is the type of animal that they are observing.
There are a lot of animals in the ocean, therefore to have a scope of their
study of observation, they are controlling or limiting their observations on
the whale alone. Hence the controlled parameter is the type of animals, whales.</span>
