<span>Answer:
To determine hybridization you count the # of un-bonded PAIR of electrons. Theny you count bonded domains (a double bond still counts as one bonded domain, so does a triple bond).
For the first Carbon on top, you see it bonded to 2 oxygens and 1 carbon.However, when you count up the valence electrons that would be present, there is supposed to be 2 more electrons (ONE electron pair) on carbon. To do hybridization you must also be familiar with drawing lewis structures. So when you draw the pair of un-bonded electrons on carbon, you see that there is ONE un-bonded electron pair, and THREE bonded domains.
ONE plus THREE = FOUR.
so that would be sp3.</span>
Answer:
expanded use of fossil fuels
Explanation:
In cellular biology, active transport is the movement of molecules across a membrane from a region of their lower concentration to a region of their higher concentration—against the concentration gradient. Active transport requires cellular energy to achieve this movement. There are two types of active transport: primary active transport that uses ATP, and secondary active transport that uses an electrochemical gradient. An example of active transport in human physiology is the uptake of glucose in the intestines.
The answer is 1296.75 g
Molar mass of <span>Fe2(SO4)3 is the sum of atomic masses of its elements:
Mr(</span>Fe2(SO4)3) = 2Ar(Fe) + 3Ar(S) + 12Ar(O)
Ar(Fe) = 55.84 g/mol
Ar(S) = 32.06 g/mol
Ar(O) = 16 g/mol
Mr(Fe2(SO4)3) = 2 * 55.84 + 3 * 32.06 + 12 * 16
= 111.68 + 96.18 + 192
= 399.9 g/mol
Thus there are 399.9 g in 1 mol and x grams in 3.25 mol:
399 g : 1 mol = x : 3.25 mol
x = 399 g * 3.25 mol : 1 mol
x = 1296.75 g
Endocytosis is a type of bulk transport which involves the movement of large particles through the membrane in and out