Answer: Bubbles formed would be evidence that a chemical reaction took place when the solutions were combined.
Explanation:
As it is given that both 
and HCl chemically combine together leading to the formation of NaCl, water and carbon dioxide gas. As the gas is forming and its formation will also form bubbles into the solution.
This formation of bubbles actually indicate that a chemical reaction has taken place. As molecules of a gas are held by Vander waal forces so, this gas will readily escape into the atmosphere.
Thus, we can conclude that bubbles formed would be evidence that a chemical reaction took place when the solutions were combined.
Answer:
42 L
Explanation:
de los parámetros en la pregunta;
V1 = 358L
T1 = 152 ° C + 273 = 425 K
P1 = 470 mmHg × 1 atm / 760 mmHg = 0.6atm
V2 =?
P2 = 6 atmósferas
T2 = 500 K
P1V1 / T1 = P2V2 / T2
P1V1T2 = P2V2T1
V2 = P1V1T2 / P2T1
V2 = 0,6 × 358 × 500/6 × 425
V2 = 107400/2550
V2 = 42 L
Answer: Concentration of N₂ is 4.8.
M.
Explanation: 
is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For
N2(g) + 2 O2(g) ⇄ 2 NO2(g)
= ![\frac{[NO2]^{2} }{[N2][O2]^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNO2%5D%5E%7B2%7D%20%7D%7B%5BN2%5D%5BO2%5D%5E%7B2%7D%20%7D)
From the question concentration of NO2 is twice of O2:
[NO2] = 2[O2]
Substituting this into :
= ![\frac{[2O2]^{2} }{[N2][O2]^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2O2%5D%5E%7B2%7D%20%7D%7B%5BN2%5D%5BO2%5D%5E%7B2%7D%20%7D)
8.3.
= ![\frac{4O2^{2} }{[N2].O2^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B4O2%5E%7B2%7D%20%7D%7B%5BN2%5D.O2%5E%7B2%7D%20%7D)
[N2] = 
[N2] = 
[N2] = 4.8.
The concentration of N2 in the equilibrium is [N2] = 4.8.M.
The three particles found in an atom are the protons, neutrons and electrons. Protons have a positive charge. Electrons have negative charge. Lastly, neutrons have no net electrical charge. Protons and neutrons are much heavier than electrons and are located in the center of the atom.
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.
