First find the volume of the cubic lattice given edge
length of 646 pm or 6.46x10^-8 cm.
volume = (6.46x10^-8 cm)^3
volume = 2.7x10^-22 cm^3
The total mass of the lattice is:
mass = (5.850 g/cm^3) * 2.7x10^-22 cm^3
mass = 1.577x10^-21 grams
The molar mass of tin is 118.71 g/mol and the Avogadros
number is 6.022 x 10^23 atoms/mol, hence:
Sn atoms = [1.577x10^-21 g / (118.71 g/mol)] * 6.022 x
10^23 atoms/mol
Sn atoms = 8 atoms
Answer: final temperatures will be
a) water 21 C
b) concrete 20.005 C
c) steel 20.008 C
d) mercury 53 C
Explanation:
Change in temp dT = dH / (mass x specific heat)
Specific heat of these materials can be found from many sources:
water = 1 kcal / kg C
concrete = 210 kcal / kg C
steel = 114 kcal / kg C
mercury = 0.03 kcal /kg C
So dT (water) from 1 kcal heat into 1 kg water = 1 kcal / (1 kg x 1 kcal/kg C) = 1 C therefore the final temperature is 20 + 1 = 21 C
But dT (steel) = 1 kcal / (1kg x 114 kcal/kg C) = 0.008 C so the final temperature is 20 + 0.008 = 20.008 C
The results for concrete and mercury are calculated in the same way
Answer:
A. Methanol
B. 2-chloropropan-1-ol
C. 2,2-dichloroethanol
D. 2,2-difluoropropan-1-ol
Explanation:
Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.
This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).
By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.
By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.
Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.
Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.
From above,
A. Methanol
B. 2-chloropropan-1-ol
C. 2,2-dichloroethanol
D. 2,2-difluoropropan-1-ol
D) because both reactions are occurring at the same rate. They are not equal but their concentrations are constant.
A complex, ML₆²⁺, is violet. The same metal forms a complex with another ligand, Q, that creates a weaker field. MQ₆²⁺, be expected to show green color.
<h3>What is spectrochemical series?</h3>
The ligands (attachment to a metal ion) are listed in the spectrochemical series according to the strength of their field. The series has been created by superimposing several sequences discovered through spectroscopic research because it is impossible to produce the full series by examining complexes with a single metal ion. The halides are referred to be weak-field ligands whereas the ligands cyanide and CO are strong-field ligands. Medium field effects are claimed to be produced by ligands like water and ammonia.
To know more about the spectrochemical series, visit:
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