Answer:
3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂
Explanation:
3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂
In this type of reaction, one substance is replacing another:
A + BC → AC + B
In a single displacement reaction, atoms replace one another based on the activity series. Elements that are higher in the activity series. Also, if the element that is to replace the other in a compound is more reactive the reaction will occur. If it is less reactive, there will be no reation.
In the first equation, fluorine is more reactive than bromine. Therefore, bromine cannot replace bromine.
In the second equation, the displacement is between hydrogen and aluminium. Hydrogen is lower in the activity series, this implies that aluminum will replace it.
Answer:
False
Explanation:
That's because elements in a compound combine and become an entirely different substance with its own unique properties.
A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
Answer:
the difference is tyat eruptions of less gassy and more gassy is that the less gassy doesnt retain as much gas as the more gassy one and thus the eruption of the less gassy is less damage to the more gassy
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
