Answer:
It is necessary because Trans-cinnamic is a limiting reagent in the mixture reaction while Bromine is the excess reagent
Explanation:
It is necessary to maintain excess bromine in the reaction mixture because Bromine is the excess reagent in the reaction mixture and if it's quantity is less it would consume the limiting reagent ( Trans-cinnamic ) completely . hence Bromine should maintain excess quantity in the reaction mixture
Given:
K = 0.71 = Kp
The reaction of sulphur with oxygen is
S(s) + O2(g) ---> SO2(g)
initial Pressure 6.90 0
Change -x +x
Equilibrium 6.90-x x
Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)
4.899 - 0.71x = x
4.899 = 1.71x
x = 2.86 atm = pressure of SO2 formed
temperature = 950 C = 950 + 273.15 K = 1223.15 K
Volume = 50 L
Let us calculate moles of SO2 formed using ideal gas equation as
PV = nRT
R = gas constant = 0.0821 L atm / mol K
putting other values
n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles
Moles of Sulphur required = 1.42 moles
Mass of sulphur required or consumed = moles X atomic mass of sulphur
mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur
A. SO2Cl2(g) --> SO2(g) + Cl2(g)
<span>1 mole of SOCl2 becomes 1 mole SO2 and 1 mole Cl2 </span>
<span>1 mole --> 2 moles </span>
<span>entropy increases </span>
Answer:
116.88g of table salt (NaCl) contains two formula units
Explanation:
Now,
We know that 1 formula unit of sodium chloride has a molar mass of 58.44g/mol
Hence;
Mass of 1 formula unit = 58.44g
Mass of x formula units = 116.88g
x = 116.88g * 1 formula unit/58.44g
x = 2 formula units
Therefore;
116.88g of table salt (NaCl) contains two formula units
<span>STP means standard temperature
and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown
gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the
numer of moles and can be represented as mass of the gas, m, divided by the
molar mass, c. so we have,</span>
PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P
substitute the values into the equation,
c = [(0.63g/L)(0.08206
L-atm/mol-K)(273K)]/(1atm)
<u>c = 14.11 g/mol</u>
