What type of hybridization is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen?: and
1 answer:
<u></u><u> hybridization and 1 lone pair</u> is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen.
b. hybridization and 1 lone pair
<u>Explanation:</u>
The Nitrogen particle is hybridized with one crossover orbital involved by the solitary pair. Likewise, nitrogen is
hybridized which implies that it has four
half and half orbitals. The sub-atomic structure of water is predictable with a tetrahedral game plan of two solitary sets and two holding sets of electrons. Two of the
hybridized orbitals cover with s orbitals from hydrogens to frame the two N-H sigma bonds.
Nitrogen utilizes orbitals to accomplish this geometry. Three of the mixtures are utilized to frame bonds to hydrogen and the fourth contains the solitary pair. Whereas lone pairs are the pairs of electron on atoms that don't take an interest in the holding bonding of two atoms. To distinguish solitary matches in a particle, make sense of the number of valence electrons of the molecule and subtract the number of electrons that have partaken in the holding.
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Answer:
1. a
b
c
2.
Explanation:
a). The work done by the tension is:
b). The work done potential of gravity
c). The work done by the normal force
2. The increase in thermal energy is:
Answer:
Explanation:
A simple pendulum is a system consisting of a mass attached to a string, and oscillating in a periodic motion, back and forth, along an equilibrium position.
The period of a pendulum is the time it takes for the pendulum to complete one oscillation.
The period of a pendulum is given by the equation
where
L is the length of the pendulum
g is the acceleration due to gravity
From the formula, we see that the period of a pendulum does not depend on the mass.
Therefore, the only 2 factors affecting the period of a pendulum are:
- The length of the pendulum: the longer it is, the longer the period of oscillation
- The acceleration due to gravity: the greater it is, the shorter the period of the pendulum
The concept of this question can be well understood by listing out the parameters given.
- The mass of the block = 51 g = 51 × 10⁻³ kg
- The distance of the block from the table = 30 cm
- Length of the spring = 30 cm
The purpose is to determine the spring constant.
Let us assume that the two blocks are Block A and Block B.
At point A on block A, the initial velocity on the block is zero
i.e. u = 0
We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.
From the above formula,
The distance (S) = 30 cm; we need to convert the unit to meter (m).
- Since 1 cm = 0.01 m
- Then, 30cm = 0.3 m
The acceleration (g) due to gravity = 9.8 m/s²
∴
inputting the values into the equation above, we have;
By dividing both sides by 4.9, we have:
However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.
By applying the equation of the time period of a simple harmonic motion.
where the relation between time (t) and period (T) is:
T = 2t
T = 2(0.247)
T = 0.494 seconds
By making the spring constant k the subject of the formula:
Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.
Learn more about simple harmonic motion here:
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