Answer : 1721.72 g/qt are in 18.2 g/cL
Explanation :
As we are given: 18.2 g/cL
Now we have to convert 18.2 g/cL to g/qt.
Conversions used are:
(1) 1 L = 100 cL
(2) 1 L = 1000 mL
(3) 1 qt = 946 qt
The conversion expression will be:


Therefore, 1721.72 g/qt are in 18.2 g/cL
Answer:
Explanation:
The following equation relates to Born-Haber cycle

Where
is enthalpy of formation
S is enthalpy of sublimation
B is bond enthalpy
is ionisation enthalpy of metal
is electron affinity of non metal atom
is lattice energy
Substituting the given values we have
-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L
= - 707 KJ / mol
There is this thing called isotopes
it means that atoms of the same element can have a different number of neutrons.
if there is a change in the no. of neutrons, there will definitely be a change in the mass number.
so the answer is A) mass number
Answer:
Explanation:
In KCL, the elements to combine to form KCl are K and Cl.
Potassium (K) is an element of group 1 and they're known to have 1 valence electron in its outer most orbital. This electron is donated and used for bonding. It is every element desirability to attain octet or duet electronic configuration I.e by having 2 electrons or 8 electrons in its outer orbital. Potassium has a lower electro negativity and higher ionization energy, hence it is easier to donate electron than to recieve it. Meanwhile, Chlorine (Cl) is an element of group 17 which are characterized for having 7 electrons in its outermost orbital requiring just one electron to form an octet configuration, chlorine has a higher electro negativity and lower ionization energy hence it is easier to accept electron than to donate it.
The nature of bond formed from K and Cl to give KCl is called electrovalent bonds.
Equation of reaction
K + Cl → KCl
Answer:
Here's what I get
Explanation:
1. Balanced equation
HQ⁻ + CH₃-Br ⟶ HQ-CH₃ + Br⁻
(I must use HQ because the Brainly Editor thinks the O makes a forbidden word)
2. Mechanism
HQ⁻ + CH₃-Br ⟶[HQ···CH₃···Br]⁻⟶ HQ-CH₃ + Br⁻
A C B
The hydroxide ion attacks the back side of the carbon atom in the bromomethane (A).
At the same time as the Q-H bond starts to form, the C-Br bond starts to break.
At the half-way point, we have a high-energy intermediate (C) with partially formed C-O and C-Br bonds.
As the reaction proceeds further, the Br atom drops off to form the products — methanol and bromide ion (B).
3. Energy diagram
See the diagram below.