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℃.
A catalyst will ALWAYS increase the reaction rate so C
Molarity after dilution : 0.0058 M
<h3>Further explanation
</h3>
The number of moles before and after dilution is the same
The dilution formula
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
M₁=0.1 M
V₁=6.11
V₂=105.12
The answer is C I got it right so hope it helped ;)
Answer:
Mass = 90.28 g
Explanation:
Given data:
Mass of Ca(OH)₂ = ?
Volume of solution= 1.5 L
Molarity of solution = 0.81 M
Solution:
First of all we will calculate number of moles.
Molarity = number of moles / volume in L
by putting values,
0.81 M = Number of moles / 1.5 L
Number of moles = 0.81 M × 1.5 L
Number of moles = 1.22 mol
Mass of Ca(OH)₂ in gram:
Mass = number of moles × molar mass
Mass = 1.22 mol × 74.09 g/mol
Mass = 90.28 g
