The color emitted be larger atoms is lower in energy then the light emitted by smaller atoms
Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
In an ionic compound the atoms are linked via ionic bonds. These are formed by the transfer of electrons from one atom to the other. The atom that loses electrons gains a positive charge whereas the atom that accepts electrons gains a negative. This happens in accordance with the octet rule wherein each atom is surrounded by 8 electrons
In the given example:
The valence electron configuration of Iodine (I) = 5s²5p⁵
It needs only one electron to complete its octet.
In the given options:
K = 4s¹
C = 2s²2p²
Cl = 3s²3p⁵
P = 3s²3p³
Thus K can donate its valence electron to Iodine. As a result K, will gain a stable noble gas configuration of argon while iodine would gain an octet. This would also balance the charges as K⁺I⁻ creating a neutral molecule.
Ans: Potassium (K)
Answer:
The pH of the solution is 7.00 .
Explanation:
The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:
pH = -log (H30 +)
pH= -log (1x10^-7)
<em>pH= 7.00</em>
Answer : B) In step 2, there was a chemical change which was observed in sugar.
Explanation : In A step 2 there was a physical change that was seen. By just boiling the dissolved salt solution salt was obtained. Therefore, it is a physical change. In B step 2 there was a chemical change seen as sugar solution was thickened and turned brown. It was not obtained in its original form; there was a chemical reaction that took place during sugar evaporation. As chemical change is the one where the reaction is irreversible.
Therefore only in B step 2 there was a chemical change that was observed.
