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
The answers that fit the blanks are SMALL and LITTLE, respectively. The particles or molecules or fas are small which makes it loose and easily moves around, and these only exert little attraction for other gas particles. The answer for this would be option D.
The atomic mass number does<span> not change because a </span>beta<span> particle has a much smaller </span>mass<span> than the </span>atom<span>. The </span>atomic number<span> goes up because a neutron has turned into an extra proton. </span>Beta decay<span> is fundamentally different from alpha </span>decay<span>. An alpha particle is made of two protons and two neutrons.</span>
Answer:
42.38875878%
Explanation:
i divided 4.27 g from 1.81g using a percentage calculator but im not sure if its correct
<h2>Answer:</h2>
<em>8.67kJ/mol</em>
<h2>Explanations</h2>
The formula for calculating the amount of heat absorbed by the water is given as:

Determine the moles of KI

Since heat is lost, hence the enthalpy change of the solution will be negative that is:

Determine the enthalpy of solution in kJ•mol-1

Hence the enthalpy of solution in kJ•mol-1 for KI is 8.67kJ/mol