This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
<u>Explanation</u>:
The standard enthalpy change of reaction, Δ
H
∘
, is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.
C
(s] + O
2(g]
→
CO
2(g]
Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.
First to convert grams of carbon into moles,
use carbon's molar mass(12.011 g).
Moles of C = mass in gram / molar mass
= 0.327 g / 12.011 g
Moles of C = 0.027 moles
Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.
= 0.027 moles C 
393 kJ
Heat released = 10.611 kJ.
So, when 0.027 moles of carbon react with enough oxygen gas, the reaction will give off 10.611 kJ of heat.
This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
The answer is: lose electrons and form positive ions.
Most metals have strong metallic bond, because of strong electrostatic attractive force between valence electrons (metals usually have low ionization energy and lose electrons easy) and positively charged metal ions.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
For example, magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.
Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.
Answer:
Glucose also called(Energy)
Answer:
C.) HOCl Ka=3.5x10^-8
Explanation:
In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below
we Know that
pKa= -log(Ka)
therefore
A) pKa of HClO2 = -log(1.2 x 10^-2)
=1.9208
B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644
C) pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45
D) pKa of HCN = -log(4 x 1 0^-10)= 9.3979
If we consider the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution
The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.
So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.
Hence, HOCl will be chosen for buffer construction.
