Answer:
The reason for the suspicion was because the manner in which iodine reacted chemically as well as its other chemical properties, indicated that it belonged in the same group as chlorine and bromine, while the much heavier tellurium should be placed in the previous group
The suspicion was proved to be correct when the atomic number of tellurium was found to be 52 and that of iodine was found to be 53 by later scientists
Explanation:
<span>100.
ppb of chcl3 in drinking water means 100 g of CHCl3 in 1,000,0000,000 g of water
Molarity, M
M = number of moles of solute / volume of solution in liters
number of moles of solute = mass of CHCl3 / molar mass of CHCl3
molar mass of CHCl3 = 119.37 g/mol
number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol
using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters
M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer
Molality, m
m = number of moles of solute / kg of solvent
number of moles of solute = 0.838
kg of solvent = kg of water = 1,000,000 kg
m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer
mole fraction of solute, X solute
X solute = number of moles of solute / number of moles of solution
number of moles of solute = 0.838
number of moles of solution = number of moles of solute + number of moles of solvent
number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles
number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles
X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer
mass percent, %
% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =
% = 10 ^ - 6 % <------- answer
</span>
Answer:
d , before the molten rock becomes lava, it is first magma, and most people know that lava is ejected from volcanoes
Answer:
3.24 × 10^5 J/mol
Explanation:
The activation energy of this reaction can be calculated using the equation:
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
Where; Ea = the activation energy (J/mol)
R = the ideal gas constant = 8.3145 J/Kmol
T1 and T2 = absolute temperatures (K)
k1 and k2 = the reaction rate constants at respective temperature
First, we need to convert the temperatures in °C to K
T(K) = T(°C) + 273.15
T1 = 325°C + 273.15
T1 = 598.15K
T2 = 407°C + 273.15
T2 = 680.15K
Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)
ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4
7.831 = Ea(2.417 × 10^-5)
Ea = 3.24 × 10^5 J/mol
Answer:
i) 0.5071 (kg/s)
ii) -1407.1 kj/kg
iii) 204.05 Kw
iv) 5.881
v) 9.238
Explanation:
Given Data:
evaporation temperature ( T ) = 4°c = 277.15 K
Condensation Temperature ( T ) = 34°c = 307.15 K
<em>n</em> ( compressor efficiency ) = 0.76
refrigeration rate = 1200 kJ.s^-1
i) determine the circulation rate of the refrigerant
m =
=
------- 1
Q = 1200 Kj.s^-1
H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )
H4 = entropy at step 4 = 142.4 ( kJ/ kg )
back to equation 1
m ( circulation rate of refrigerant ) = 0.5071 (kg/s)
ii) heat transfer rate in the condenser
Q = m ( H4 - H3 )
= 0.5071 ( 142.4 - 2911.27 )
= -1407.1 kj/kg
where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )
iii) power requirement
w = m * ΔH23
= 0.5071 (kg/s) * 402.37 (kj/kg) = 204.05 Kw
where: ΔH23 =
=
= 402.37 (kj/kg)
iv) coefficient of performance of a cycle
W = Qc / w
= 1200 Kj.s^-1/ 204.05 kw
= 5.881
v) coefficient of performance of a Carnot refrigeration cycle

= 277.15 / ( 307.15 - 277.15 )
= 9.238