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.
Answer:
glycerol
Explanation:
it is thick sweet liquid.It is type of alcohol which has 3 OH groups
Answer:
An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements. In general, the size of an atom will decrease as you move from left to the right of a certain period.
Explanation:
Hello:
In this case, we will use the Clapeyron equation:
P = ?
n = 8 moles
T = 250 K
R = 0.082 atm.L/mol.K
V = 6 L
Therefore:
P * V = n * R * T
P * 6 = 8 * 0.082* 250
P* 6 = 164
P = 164 / 6
P = 27.33 atm
Hope that helps!
Answer:
The correct options are;
C. The magnitude of attraction from its nucleus
D. The distance between the electrons and its nucleus
Explanation:
The atomic radius reduces, within a given period, as we move from left to right, the number of protons increases alongside the number of electrons and the while the quantum shell to which the extra electrons are added to is the same. Therefore, the radius of the atom is dependent on the magnitude of the attraction from the nucleus
Similarly, as we progress to the next period, with an extra quantum shell, the atomic radius is seen to increase.
Therefore, the atomic radius is determined by the distance between the electrons and its nucleus.