Answer:
The answer to your question is 7160 cm
Explanation:
Data
diameter = 1 mm
length = ?
amount of gold = 1 mol
density = 17 g/cm³
Process
1.- Get the atomic mass of gold
Atomic mass = 197 g
then, 197g ------------ 1 mol
2.- Calculate the volume of this wire
density = mass/volume
volume = mass/density
volume = 197/17
volume = 5.7 cm³
3.- Calculate the length of the wire
Volume = πr²h
solve for h
h = volume /πr²
radius = 0.05 cm
substitution
h = 5.7/(3.14 x 0.05²)
h = 5.7 / 0.0025
h = 7159.2 cm ≈ 7160 cm
Electrons in sigma <span>bonds remain localized between two atoms. Sigma </span><span>bond results from the formation of </span><span>a molecular orbital </span><span>by the end to </span><span>end overlap of atomic </span>orbitals. Electrons<span> in pi</span> bonds can become delocalized between more than two atoms. Pi bonds result from the formation of molecular orbital by side to side overlap of atomic orbitals.
<span> </span>
Answer:
(A) pH < 1 the predominant form is the cation: H3C-C(H)(NH3+)-COOH
(B) pH = pl the predominant form is the zwitterion H3C-C(H)(NH3+)-COO-
(C) pH > 11 the predominant form is the anion: H3C-C(H)(NH2)-COO-
(D) Does not occurs in any significant pH: H3C-C(H)(NH2)-COOH
Explanation:
Amino acids are bifunctional because they have an amine group and a carboxyl group. The amine group is a weak base and the carboxyl group is a weak acid, but the pKa of both groups will depend on the whole structure of the amino acid. Also, every amino acid has an isoelectric point (pI), which means the pH were the predominant form of the amino acid is the zwitterion. The structure of the alanine (CH3CH2NH2COOH) shows it has the carboxyl group at C1 with a pKa1 of 2.3 and the amino group at C2 whit the pKa2 of 9.7. The isoelectric poin (pI) of Alanine is 6. Consequently, the protonation of the molecule will depend on the pH of the solution. There are three possibilities:
1) If the pH is under the pKa of the carboxyl group (2.3) the predominant form will be with the amino group protonated, forming a cation (CH3CH(NH3+)COOH).
2) If the pH is between pKa1 (2.3) and pKa2 (9.7) the predominant form will be the zwitterion (CH3CH(NH3+)(COO-)).
3) If the pH is upper the pKa2 of the amino group (9.7) the predominant form will be with the carboxyl group deprotonated, forming an anion (CH3CHNH2(COO-)).
Concept:
<em><u>Latent Heat of Vaporization</u></em>: It is defined as the amount of heat required to change the state of mater without changing of its temperature.
From the given question, the temperature at the boiling point remained constant despite the continued addition of heat by the Bunsen burner. <em>Actually,</em> this amount of heat is used by water to break the intermolecular bonds between the water molecules in the form of latent heat that converts the liquid state of water into vapor state of water.
Hence, the correct option will be d.<u>The energy was used to break the intermolecular bonds between the water molecules. </u>
<span>Answer:
Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass.
râšM = constant
Therefore for two gases the ratio rates is given by:
r1 / r2 = âš(M2 / M1)
For Cl2 and F2:
r(Cl2) / r(F2) = âš{(37.9968)/(70.906)}
= 0.732 (to 3.s.f.)</span>