<span>Zn⁰ + 2H⁺ ------> Zn²⁺ + H2⁰
H⁺ ion has oxidation number +1.
Zn²⁺ ion has oxidation number +2.
Atom of Zn has electric charge 0, and each hydrogen atom in the molecule H2 have oxidation number 0. So, </span> Zn and each hydrogen atom in H2 have oxidation numbers equals "0".<span>
</span><span>Answer is
D. Zn and each hydrogen atom in H2</span><span>
</span>
Answer:
Colloids (heterogeneous)
The difference between a colloid and a suspension is that the particles will not settle to the bottom over a period of time, they will stay suspended or float. An example of a colloid is milk. Milk is a mixture of liquid butterfat globules dispersed and suspended in water.
The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.
<h3>
What is relative abundance?</h3>
- The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
- When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
- Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.
To learn more about relative abundance with the given link
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The balanced equation for the reaction is as follows;
Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O
stoichiometry of Ca(OH)₂ to HBr is 1:2
number of Ca(OH)₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol
Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol
1 mol of Ca(OH)₂ needs 2 mol of HBr for neutralisation
therefore 0.010 mol of Ca(OH)₂ needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised
but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol
then pH of the medium can be calculated using the excess H⁺ ions
HBr is a strong acid therefore complete ionization
[HBr] = [H⁺]
[H⁺] = 0.020 mol / (100.0 + 400.0 mL)
= 0.020 mol / 0.5 L
= 0.040 mol/L
pH = -log[H⁺]
pH = - log [0.040 M]
pH = 1.40
pH of the medium is 1.40
