Triangles in an octagon
1 answer:
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Answer:
The root mean squared velocity for CF2Cl2 is
Explanation:
From the question we are told that
The temperature is
Root Mean Square velocity is mathematically represented as
Where T is the temperature
MW is the molecular weight of gas
R is the gas constant with a value of
For CF2Cl2 its molecular weight is 0.121 kg/mol
Substituting values
Answer : The correct option is, Subscript numbers tell us how many of the element to their upper-left there are in the molecule.
Explanation :
Subscript number : It indicates that the total number of atoms of an element present in a molecule.
While witting a chemical formula, first we have to write the symbol of an element and then write the number in the bottom of the symbol.
For example :
is the chemical formula of the glucose molecule. The glucose molecule is made up of six carbon atoms, twelve hydrogen atoms and 6 oxygen atoms.
The number 6 of carbon atom is written after and in the bottom (lower-right) of the symbol of carbon element (C). Similarly, the number 12 is written lower-right of the symbol of hydrogen element (H) and the number 6 is written lower-right of the symbol of oxygen element (O).
Therefore, the subscript numbers tell us how many of the element to their upper-left there are in the molecule.
Answer:
An example of oxygen–hemoglobin (O2–Hb) dissociation curves from (A) one penguin at pH 7.5, 7.4 and 7.3, and (B) the emperor penguin, the bar-headed goose (Anser indicus) (Black and Tenney, 1980) and the domestic duck (Anas platyrhynchos, forma domestica) (Hudson and Jones, 1986) at pH 7.4. Note that as for the bar-headed goose, the O2–Hb dissociation curve of the emperor penguin is significantly left-shifted as compared with the domestic duck (and most birds). The bar-headed goose photo is courtesy of Graham Scott; the domestic duck photo is by Maren Winter (licensed under the terms of the GNU Free Documentation License, Version 1.2 or any later version); the penguin photo is by J.M.
Explanation:
The resulting regression equations from the plots of log[SO2/(100–SO2)] vs log(PO2) (all saturation points, all penguins combined) were:
pH 7.5: log[SO2/(100–SO2)] = 2.92589 × log(PO2) – 4.24338 (N=43, r2=0.98, P<0.0001),
pH 7.4: log[SO2/(100–SO2)] = 2.94767 × log(PO2) – 4.39858 (N=70, r2=0.98, P<0.0001),
pH 7.3: log[SO2/(100–SO2)] = 3.04945 × log(PO2) – 4.72019 (N=38, r2=0.99, P<0.0001),
pH 7.2: log[SO2/(100–SO2)] = 3.15958 × log(PO2) – 4.97618 (N=9, r2=0.99, P<0.0001).
