Blank 1: polar
The difference in electronegativity between N and H causes electrons to preferentially orbit N, making the bond polar.
Blank 2: trigonal pyramidal
There are four “things” attached to N - 3 H’s and 1 lone pair of electrons. The four things together are arranged into a tetrahedral formation. However, the lone pairs don’t actually contribute to the shape of the molecule per se; it’s only the actual atoms that do. The lone pair creates a bit of repulsion that pushes the 3 H’s down, creating a trigonal pyramidal shape (as opposed to a trigonal planar one).
Blank 3: polar
The molecule as a whole is also polar because the “things” around it, though arranged in a tetrahedral pattern, are not all the same. The side of the molecule with the lone pair is slightly negative, while the side with the 3 H’s is slightly positive due to the differences in electronegativity described above.
Answer:
The particles in the neutral paper can shift, causing the paper to become polarized and attracted to the rod.
Explanation:
The neutral paper has an even distribution of its electrons throughout the paper. If a charged rod is brought near the neutral paper, this can cause the electrons in the paper to shift. If the rod is negative, the electrons will be repelled from the rod and cause the molecules in the paper to have a slight positive charge on the part of the paper closest to the rod. If the rod is positive, the electrons will be attracted to the rod and cause a slight negative charge on the side of the paper closest to the rod.
Answer:
It is called tempering. Its tensile strength may reduce but it will become more rigid and hard to break compared to the original metal.
Explanation:
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Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
A model is developed for predicting oxygen uptake, muscle blood flow, and blood chemistry changes under exercise conditions. In this model, the working muscle mass system is analyzed. The conservation of matter principle is applied to the oxygen in a unit mass of working muscle under transient exercise conditions. This principle is used to relate the inflow of oxygen carried with the blood to the outflow carried with blood, the rate of change of oxygen stored in the muscle myoglobin, and the uptake by the muscle. Standard blood chemistry relations are incorporated to evaluate venous levels of oxygen, pH, and carbon dioxide.
Explanation:
