Answer:
17 protons, 20 neutrons, and 17 electrons.
Explanation:
A periodic table can be defined as the standard arrangement of chemical elements by atomic number, electronic configuration and chemical properties in a tabular form.
Generally, a proper representation of the mass number and atomic number of chemical elements is key and very important in chemistry.
Furthermore, as a rule, it should be noted that the mass number (nucleon number) is always larger than the atomic number(number of proton).
The mass number of this neutral atom of Cl-37 is 37 and we know that the atomic number (number of protons) of chlorine is 17. Also, the atomic number of an element is equal to the number of its electrons.
A neutral atom of Cl-37 has 17 protons, 20 neutrons, and 17 electrons.
Hence, a neutral atom of Cl-37 can be identified based on its number of protons because it represent its atomic number, which is what is used to differentiate an atom of an element from the atom of another chemical element.
Answer:
ΔpH = 0.20
Explanation:
The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2
HCO₃⁻ ⇄ H⁺ + CO₃²⁻
There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.
Using Henderson-Hasselbalch formula:
pH = pka + log [Base] / [Acid]
pH = 10.2 + log 0.342mol / 0.479mol
<em>pH = 10.05</em>
NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:
NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺
0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:
CO₃²⁻: 0.342mol + 0.091mol: 0.433mol
HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol
Using Henderson-Hasselbalch formula:
pH = pka + log [Base] / [Acid]
pH = 10.2 + log 0.433mol / 0.388mol
pH = 10.25
That means change of pH, ΔpH is:
ΔpH = 10.25 - 10.05 = <em>0.20</em>
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I hope it helps!
M=D*V,
V=m/D
m=15 g
D=3 g/ml
V=15 g/3 g/ml=5 ml
<span>2C2H6 + 7O2 = 4CO2 + 6H2O
</span>
According to the equation of the reaction of ethane combustion, ethane and carbon dioxide have following stoichiometric ratio:
n(C2H6) : n(CO2) = 1 : 2
n(CO2) = 2 x n(C2H6)
n(CO2) = 2 x 5.2 = 10.4 mole of CO2 is formed
A) sodium fluoride
B) rubidium oxide
C) boron trichloride
D) dihydrogen selenide
E) tetraphosphate hexoxide
F) iodine trichloride
