Karst topography can cause drinking water to become more polluted
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:
Potential energy
Explanation:
Potential energy is the energy possed by an object at rest. It is otherwise referred to as the stored energy due position.
Traditionally they include boron from group 3A, silicon and germanium in group 4A, aresnic and antimony in group 5A and tellurium from group 6A, although sometimes selenium, astatine, polonium and even bismuth have also been considered as metalloids. Typically metalloids are brittle and show a semi-metallic luster.
The six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. Five elements are less frequently so classified: carbon, aluminium, selenium, polonium, and astatine.
First convert celcius to Kelvin.
20 + 273 = 293K
31 + 273 = 304K
Now we can set up an equation based on the information we have.
V1 = 5
P1 = 365
T1 = 293
V2 = 5
P1 = x
T2 = 304
The equation be: 
Now just solve.
1825/293 = 5x/304
Cross multiply.
554800 = 1465x
Divide both sides by 1465
x = 378.7030717 which can then be rounded to 378.7 mmHg