B: 2
If it's right plz give brainliest lol <3
Answer:
Purpose: To become familiar with the techniques for separation of amixture of solids.
Explanation:
a mixture of pure substances. If you have a mixture of tennis ballsand marbles (not pure substances by the way), it would be easy toseparate the mixture. However, it is more difficult to separate asand (also not a pure substance) and salt mixture. Even with verygood tweezers and a magnifying glass, it would be extremelytedious. You could take advantage of the fact that salt dissolvesin water and sand does not. To separate iron powder from an ironand sand mixture you can take advantage of the magnetic propertiesof iron and separate the mixture.
To summarize a complete procedure for separating a mixture ofseveral substances, it is best to prepare a flow chart. A flowchartis a schematic representation of an algorithm or a stepwiseprocess, showing the steps as boxes of various kinds, and theirorder by connecting these with arrows. Flowcharts are used indesigning or documenting a process.
Answer:
1) Increasing temperature
2) Stirring
3) Increasing surface area of salt by grinding it
M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C
m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C
T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
Since
21.2 g H2O was produced, the amount of oxygen that reacted can be obtained
using stoichiometry. The balanced equation was given: 2H₂ + O₂ → 2H₂O and
the molar masses of the relevant species are also listed below. Thus, the
following equation is used to determine the amount of oxygen consumed.
Molar mass of H2O = 18
g/mol
Molar mass of O2 = 32
g/mol
21.2 g H20 x 1 mol
H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2
<span>We then determine that
18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is
important to note that we do not need to consider the amount of H2 since we can
derive the amount of O2 from the product. Additionally, the amount of H2 is in
excess in the reaction.</span>
