Answer:
Answers are bellow.
Explanation:
The element with electron configuration 1s22s22p63s1 belong I A group in the periodic table and it is sodium because it loses one electron.
We have periodic table in attachments.
Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sodium is an alkali metal, being in group 1 of the periodic table, because it has a single electron in its outer shell, which it readily donates, creating a positively charged ion—the Na+ cation. Its only stable isotope is 23Na.
The free metal does not occur in nature, and must be prepared from compounds.
It is soft metal, reactive and with a low melting point.
Answer:
This question appears incomplete
Explanation:
The equation for the formation of iron(III) thiocyanate complex ion is
Fe³⁺ + SCN⁻ ⇄ Fe(SCN)²⁺
Three factors affect the equilibrium of a reaction
- temperature
- pressure (in gases)
- concentration
Since the formation of Fe(SCN)²⁺ is an exothermic reaction (delta H is negative); an increase in temperature will favor the backward reaction and more reactants will be produced. When the temperature is however reduced, the forward reaction will be favored and more products will be produced.
When the concentration of the reactants increases, the forward reaction is favored and hence more Fe(SCN)²⁺ is produced. However, when the concentration of Fe(SCN)²⁺ decreases, the backward reaction is favored and more reactants are produced.
Answer:
2008.3 J
Explanation:
Use the equation q=mcΔT where
q = heat
m = mass
c = specific heat capacity (for water this is 4.184 and is a given)
delta T = change in temperature in C (final - initial)
q = 30 g x 4.184 J/g*C x (48-32)
q = 2008.3 J (or 2 kJ)
The temperature at which benzene boils at a given pressure of 410 torr can be deduced from the Clausius Clapeyron equation:
ln(P2/P1) = -ΔHvap/R [1/T2-1/T1] ------(1)
where
ΔHvap = heat of vaporization of benzene
P1 = atmospheric pressure
T1 = normal boiling point
R = gas constant
These values are essentially standard values which can be obtained from any thermochemical data base
For benzene: ΔHvap = 30.72 kJ/mol and T1 = 80 C = 80 +273 = 353 K
R = 0.008314 kJ/K.mol
P1 = 760 torr
P2 = 410 torr
Substituting these values in equation (1) we get:
ln(410/760) = -30.72/0.008314 [1/T2-1/353]
T2 = 337 K
or, T2 = 337 - 273 = 64 C
