Sorry man I need some points but I do t even know the answer
Answer:
- <u>Tellurium (Te) and iodine (I) are two elements </u><em><u>next to each other that have decreasing atomic masses.</u></em>
Explanation:
The <em>atomic mass</em> of tellurium (Te) is 127.60 g/mol and the atomic mass of iodine (I) is 126.904 g/mol; so, in spite of iodine being to the right of tellurium in the periodic table (because the atomic number of iodine is bigger than the atomic number of tellurium), the atomic mass of iodine is less than the atomic mass of tellurium.
The elements are arranged in increasing order of atomic number in the periodic table.
The atomic number is equal to the number of protons and the mass number is the sum of the protons and neutrons.
The mass number, except for the mass defect, represents the atomic mass of a particular isotope. But the atomic mass of an element is the weighted average of the atomic masses of the different natural isotopes of the element.
Normally, as the atomic number increases, you find that the atomic mass increases, so most of the elements in the periodic table, which as said are arranged in icreasing atomic number order, match with increasing atomic masses. But the relative isotope abundaces of the elements can change that.
It is the case that the most common isotopes of tellurium have atomic masses 128 amu and 130 amu, whilst most common isotopes of iodine have an atomic mass 127 amu. As result, tellurium has an average atomic mass of 127.60 g/mol whilst iodine has an average atomic mass of 126.904 g/mol.
Answer:
The unknown compound is a weak acid.
Explanation:
Given that :
a 25 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution.
A buffer region was found around a pH of 3.5. We know that a pH of 3.5 is a weak acid. So, it is likely to be an organic acid
Let assume the solution of the unknown sample to be CH₃COOH
Now :
25 mL of CH₃COOH reacted with 0.115 M of NaOH
The equation for the reaction will be :
CH₃COOH + NaOH -----> CH₃COONa + H₂O
at x mole of 0.115y M of
CH₃COOH NaOH is present
If NaOH was added in excess;
CH₃COOH + NaOH -----> CH₃COONa , NaOH will be lost then CH₃COOH and CH₃COONa will be present
Therefore;
At equilibrium : Only CH₃COONa will be present but if it is above equilibrium NaOH will be present because the pH will increase due to the presence of the strong base
Answer:
c
Explanation:
it is c because that is the andwer
