Select all that apply.The requirements for one type of atom to substitute for another in a solid solution are:
All substitutions must be limited to the same element.
An atom must be identical in size.
An atom must be similar in size.
The substituting atom must be from the same period.
The substituting atom must be from the same group.
Answers
Ans: C) and E)
A solid solution is essentially a multicomponent system which is composed of a mixture of two or more elements sharing the same crystal lattice. Alloys are solid solutions.
The requirement for one type of atom to substitute for another in a solid solution are two fold
1) The atom must be similar in size
2) The substituting atom must be from the same group.
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Consider three gases: Ar, SF6, and Cl2. If 50.0 grams of these gases are placed in each of three identical containers, which container will have the highest pressure? The volume and temperature of all three containers are the same.
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The weight of oxegon
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it's core charge would be +3
Explanation:
it's core charge would be +3
Explanation:
Answers
The complete ionic equation is as follows: H+ + I
+K
+OH
H₂O
+K
+I
What is ionic equation?
Ionic equation is defined as a chemical equation which represents electrolytes in aqueous solutions and are expressed as dissociated ions. Ions present in aqueous solutions are stabilized by ion dipole interactions with the water molecules which are present.
An ionic equation can be written for any electrolyte which dissociates and reacts with the polar solvent. In a balanced ionic equation , number and type of atoms on each sides of reaction arrow are same. Even the net charge on both sides of the arrow is same.
Strong acids and bases exist as dissociated ions present in the solution and are written in forms of ions in ionic equation.Weak acids and bases do not dissociate completely and hence are written in the form of molecular formula.
Learn more about ionic equations ,here:
#SPJ2
H+(aq) + I-(aq) + K+(aq) + OH-(aq) ------> H2O(l) + K+(aq) + I-(aq)
H+(aq) + OH-(aq) ----> H2O(l)
Answers
Explanation:
As we all know that like disolves in like solvent.
Here both water and Ethanol are polar. Hence Ethanol soluble in water.
Moreover, both forms intermolecular hydrogen bonds. It enhances the solubility of ethanol.
Both water and Ethanol are called as associate liquids. Hence solubility is more.
Alcohols have higher boiling point than other hydrocarbons due to their polarity and from the fact that they form very strong intermolecular hydrogen bonding. This is due to the large difference in their electronegativity that forms between the oxygen and the hydrogen atom.
Answers
(500. mL)/(1000L) = .500L
Then convert molarity to moles:
(3.00M)(.500L) = 1.5mol NaCl
Then convert moles of NaCl to grams of NaCl by multiplying it by its molar mass:
(1.5mol NaCl)(58.44g/mol NaCl) = 87.66g NaCl are needed
Answers
Answer:
-909.3KJ/mole
Explanation:
The heat of reaction is accessible from the heat of formation of reactants and products using the formula below:
ΔH = Σ ΔHf products - Σ ΔHf reactants
Before we proceed, it is important to know that the enthalpy of formation of element is zero ,be it a single element or a molecule of an element.
From the reaction for the formation of sulphuric acid, we know we need to know the heat of formation of sulphur (vi) oxide and water. The examiner is quite generous and have us for water already.
Now we need to calculate for sulphur (vi) oxide. This is calculated as follows:
We first calculate for sulphur(iv)oxide. This can be obtained from the reaction between sulphur and oxygen. The calculation goes thus:
ΔH = Σ ΔHf products - Σ ΔHf reactants
ΔH = [ 1 mole suphur(iv) oxide × x] - [ (1 mole of elemental sulphur × 0) + (1 mole of elemental oxygen × 0]
We were already told this is equal to -296.8KJ. Hence the heat of formation of sulphur(iv) oxide is -296.8KJ.
We then proceed to the second stage.
Now, here we have 1 mole sulphur (iv) oxide reacting with 0.5 mole oxygen molecule.
We go again :
ΔH = Σ ΔHf products - Σ ΔHf reactants
ΔH = [ 1 mole of sulphur (vi) oxide × y] - [ (1 mole of sulphur (iv) oxide × -296.8) + (0.5 mole of oxygen × 0)].
We already know that the ΔH here equals -98.9KJ.
Hence, -98.9 = y + 296.8
y = -296.8KJ - 98.9KJ = -395.7KJ
We now proceed to the final part of the calculation which ironically comes first in the series of sentences.
Now, we want to calculate the standard heat of formation for sulphuric acid. From the reaction, we can see that one mole of sulphur (vi) oxide, reacted with one mole of water to yield one mole of sulphuric acid.
Mathematically, we go again :
ΔH = Σ ΔHf products - Σ ΔHf reactants
ΔH = [ 1 mole of sulphuric acid × z] - [( 1 mole of sulphur vi oxide × -395.7) + ( 1 mole of water × -285.8)].
Now, we know that the ΔH for this particular reaction is -227.8KJ
We then proceed to to open the bracket.
-227.8 = z - (-395.7 - 285.8)
-227.8 = z - ( -681.5)
-227.8 = z + 681.5
z = -227.8-681.5 = -909.3KJ
Hence, ΔH∘f for sulphuric acid is -909.3KJ/mol