How much of nacl is in 1.67 l of 0.400 m nacl? answer in units of mol.
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Answer:
Tungsten is used for this experiment
Explanation:
This is a Thermal - equilibrium situation. we can use the equation.
Loss of Heat of the Metal = Gain of Heat by the Water
Q = mΔT
Q = heat
m = mass
ΔT = T₂ - T₁
T₂ = final temperature
T₁ = Initial temperature
Cp = Specific heat capacity
<u>Metal</u>
m = 83.8 g
T₂ = 50⁰C
T₁ = 600⁰C
Cp =
<u>Water</u>
m = 75 g
T₂ = 50⁰C
T₁ = 30⁰C
Cp = 4.184 j.g⁻¹.⁰c⁻¹
⇒ - 83.8 x x (50 - 600) = 75 x 4.184 x (50 - 30)
⇒ =
j.g⁻¹.⁰c⁻¹
We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹
So metal Tungsten used in this experiment