Answer:
The International Date Line passes through the mid-Pacific Ocean and roughly follows a 180 degrees longitude north-south
line on the Earth. It is located halfway round the world from the prime meridian—the zero degrees longitude established in Greenwich
Answer:
<u>Our beaches would be unprotected</u>
In the short-term, these artificial sand hills will be destroyed by the elements. Because sand dunes protect inland areas from swells, tides, and winds, they must be protected and defended like national treasures. ... The ocean and the wind can have an unpredictable, destructive force on coastal regions.
- surfertoday
Natural sand dunes play a vital role in protecting our beaches, coastline and coastal developments from coastal hazards such as erosion, coastal flooding and storm damage. Sand dunes protect our shorelines from coastal erosion and provide shelter from the wind and sea spray.
- Waikato Regional Council
Hydroxyl ions are OH⁻ while hydronium ions are H₃O⁺ which is essentially H⁺ ions. The formula for pH is: pH = -log[H⁺]. So, the greater the concentration of H⁺ is, the lower the pH which indicates acidity. On the other hand, the greater the concentration of OH⁻, the greater the pH which indicates basicity. This is also a consequence of the equation: pH + pOH = 14.
Answer:
V₂ = 1.86 L
Explanation:
Given data:
Initial volume = 4.30 L
Initial pressure = 1 atm
Initial temperature = 273.15 K
Final temperature = 302 K
Final volume = ?
Final pressure = 2.56 atm
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂
/T₁ P₂
V₂ = 1 atm ×4.30 L × 302 K / 273.15 K × 2.56 atm
V₂ = 1298.6 atm.L.K / 699.26 K.atm
V₂ = 1.86 L
The atomic number for Pb is 82
∴ Pb has 82 protons and 206-82 = 14 protons
The actual mass of Pb nuclei is
=(82 × mass of the proton) + (124 × mass of neutron)
=(82× 1.00728) + (124 × 1.008664) amu
= 207.6713 amu
The mass of lead which is given is 205.9744 amu
∴mass defect is
m = 207.6713 - 205.9744 = 1.6969 amu
=1.6969 × 1.66054 × 10⁻²⁷kg
=2.818 × 10⁻²⁷kg
The binding energy is E = mc²
C is the speed of light in vacuum = 2.9979 × 10⁸m/s
∴ E = 2.532 × 10×⁻¹⁰ J/mol
= 2.532 × 10⁻¹⁰ × 6.023 × 10²³ J/mol
= 1.53811 × 10¹⁴ J/mol
