As we know,
1 D = 3.34 × 10⁻³⁰ C.m
So,
1.44 D = ?
Solving for 1.44 D,
= (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D
1.44 D = 4.80 × 10⁻³⁰ C.m
Dipole Moment is given as,
Dipole Moment = q × r
Solving for q,
q = Dipole Moment / r ------ (1)
Where,
Dipole Moment = 4.80 × 10⁻³⁰ C.m
r = 163 pm = 1.63 × 10⁻¹⁰ m
Putting values in eq. 1,
q = 4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m
q = 2.94 × 10⁻²⁰ C
As,
1.602 × 10⁻¹⁹ C = 1 e⁻
So,
2.94 × 10⁻²⁰ C = X e⁻
Solving for X,
X = (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C
= 0.183 e⁻
Result:
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.
Earth contains huge quantities of water in its oceans, lakes, rivers, the atmosphere, and believe it or not, in the rocks of the inner Earth. Over millions of years, much of this water is recycled between the inner Earth, the oceans and rivers, and the atmosphere. This cycling process means that freshwater is constantly made available to Earth's surface where we all live. Our planet is also very efficient at keeping this water. Water, as a vapor in our atmosphere, could potentially escape into space from Earth. But the water doesn't escape because certain regions of the atmosphere are extremely cold. (At an altitude of 15 kilometers, for example, the temperature of the atmosphere is as low as -60° Celsius!) At this frigid temperature, water forms solid crystals that fall back to Earth's surface.
Many people live faraway from freshwater sources. They need to carry their water home.
While our planet as a whole may never run out of water, it's important to remember that clean freshwater is not always available where and when humans need it. In fact, half of the world's freshwater can be found in only six countries. More than a billion people live without enough safe, clean water.
Also, every drop of water that we use continues through the water cycle. Stuff we put down the drain ends up in someone or something else's water. We can help protect the quality of our planet's freshwater by using it more wisely.
There are 4 lone pairs of electrons present in the carbon dioxide molecule
M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C
m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C
T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
Look at the letters, 3 would be MMeters
