Answer is no
The chemical reaction between iron and oxygen in the presence of moisture or water produces a new substance, which is the rust (iron oxide). In this case, it shows a chemical property.
The best answer to report is 2.900 g/ml (answer A)
explanation
The best answer to report must be in four significant figures. 2.900 g/ml is therefore the best answer to report since in is in four significant. zeros placed after other digits but behind a decimal point are significant figures.,theerfore 2.900 has four significant figures
To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
2 M x V1 = 0.1 M x .5 L
<span>V1 = 0.025 L or 25 mL of the
2 M KCl solution is needed</span>
A good example of equilibrium would be the mixing of oil and water in a closed container.
<h3>What is chemical equilibrium?</h3>
Chemical equilibrium is a condition in which the concentrations of components of a chemical reaction remain unchanged and have no tendency to change.
Of all the options, the only one where the concentrations of the component reactants cannot change is a mixture containing oil and water in a closed container.
Oil and water are immiscible and thus, their concentrations remain constant.
More on chemical equilibrium can be found here: brainly.com/question/4289021
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Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.
