1. What is the bond angle between the hydrogen atoms in an ammonia (NH3) molecule? 2. What is the bond angle between the oxygen
atoms in the carbon dioxide molecule? 3. What is the molecular geometry of a methane (CH4) molecule?
2 answers:
The answer is 107 degrees. The geometric shape for ammonia is Trigonal Pyramidal, even though its electron geometry is “Tetrahedral”. This is because ammonia has a lone pair of electrons that occupy its space like the other 3 hydrogens in the geometric structure.
The answer 180 degrees. This is because of the linear geometric structure of carbon dioxide. The oxygen atom is on either side of the carbon atom, each is bound by a double covalent bond. All the atoms are involved in the bond and there are no one pair electrons.
The answer is tetrahedral geometry. This is because all the 4 valence electrons of the carbon are involved in a bond with a hydrogen atom. The angles in a tetrahedral geometric arrangement, such as in methane, is 109.5 degrees, where the hydrogen atoms are as far apart, from each other, as possible .
You might be interested in
<h2>Answer: 0.17</h2>
Explanation:
The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":
(1)
Where:
is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing
is the Stefan-Boltzmann's constant.
is the Surface area of the body
is the effective temperature of the body (its surface absolute temperature) in Kelvin.
However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:
(2)
Where is the body's emissivity
(the value we want to find)
Isolating from (2):
(3)
Solving:
(4)
Finally:
(5) This is the body's emissivity
(a) +2.60 m/s
The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:
- a horizontal uniform motion, at constant speed
- a vertical motion, at constant acceleration (acceleration of gravity, , downward)
In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):
(b) -17.2 m/s
The vertical component of the fish's velocity instead follows the equation:
where
is the initial vertical velocity, which is zero
is the acceleration of gravity
t is the time
Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:
where the negative sign indicates the direction (downward).
(c)
The horizontal component of the fish's velocity would increase
The vertical component of the fish's velocity would stay the same.
As we said from part (a) and (b):
- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too
- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.