Answer:
37.14 %
Explanation:
Using the equation, mass, M = D1 * V1
= D2 * V2
Where,
D1 = density of the liquid Nitrogen
D2 = density of gaseous Nitrogen
V1 = volume of the liquid Nitrogen
V2 = volume of the gaseous Nitrogen
Calculating V2,
0.808 * 185 = 1.15 * V2
Volume of Nitrogen after expansion = 129.98 m3.
Volume = L * b * h
= 10 * 10 * 3.5
Volume of the room = 350 m3.
Fraction of air = volume of Nitrogen after expansion/volume of the room * 100
= 129.98/350 *100
= 37.14 %
D. Air molecules touch the warm ground, heating them up
Answer is: b. more than 7.
The endpoint is the point at which the indicator changes colour in a colourimetric titration and that is point when titration must stop.
For example, basic salt sodium acetate CH₃COONa is formed from the reaction between weak acid (in this example acetic acid CH₃COOH) and strong base (in this example sodium acetate NaOH).
Balanced chemical reaction of acetic acid and sodium hydroxide:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l).
Neutralization is is reaction in which an acid (in this example vinegar or acetic acid CH₃COOH) and a base react quantitatively with each other.
Answer:
ΔS = -661.0J/mol is the entropy change for the system
ΔS = -842J/mol.K is the entropy change for the surroundings
Explanation:
From the relationship between ΔG, T, ΔH and ΔS,
Mathematically, ΔG = ΔH - TΔS
TΔS = ΔH - ΔS
ΔS = ΔH - ΔS / T
but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)
plugging into the equation,
ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298
ΔS = -0.6610KJ/mol or in J.mol
ΔS = -661.0J/mol is the entropy change for the system
- For entropy change for the surroundings = ΔS = ΔH/T
- ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings
Answer:
Any of the six chemical elements that markup group1
of the periodic table.
Explanation:
