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:
hello your question is incomplete below is the missing part of the question
answer : 104°c
Explanation:
The Eutectic temperature for the mixture is 104°c
From the chart attached below it can be seen that the temperature from the two lines of best fit cross is 104°c
Hello Lilsavage717, I believe the answer you are looking for is C. or Amino Acids. They are mainly located in living things like plants or bugs.
Answer:
C
Explanation:
u can't touch a chemical with bare skin
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
