Explanation :
As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.
He placed the elements with similar nature in the same group.
According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.
However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.
As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127 and tellurium isotopes are tellurium-128 and tellurium-130.
Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.
The answer is boiling liquids I think
Answer:
The combustion of hydrogen–oxygen mixtures is used to produce very high temperatures (approximately 2500 °C) needed for certain types of welding operations. Consider the reaction to be
H2(g)+1/2O2(g)=H2O(g)
change in enthalpy is -241.8 kJ
What is the quantity of heat evolved, in kilojoules, when a 160 g mixture containing equal parts of H2 and O2 by mass is burned?
Explanation:
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An experimental group, also known as a treatment group, receives the treatment whose effect researchers wish to study, whereas a control group does not. They should be identical in all other ways
Answer:
6.93
Explanation:
Step 1: Given data
- Standard Gibbs free energy (∆G°): -5.20 kJ
- Equilibrium constant (K): ?
Step 2: Convert the temperature to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 50°C + 273.15
K = 323 K
Step 3: Calculate K
We will use the following expression.
∆G° = -R × T × ln K
-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K
K = 6.93
