Answer:
[KBr] = 454.5 m
Explanation:
m is a sort of concentration that indicates the moles of solute which are contianed in 1kg of solvent.
In this case, the moles of solute are 0.25 moles.
Let's determine the mass of solvent in kg.
Density of heavy water, solvent, is 1.1 g/L and our volume is 0.5L.
1.1 g = mass of solvent / 0.5L, according to density.
mass of solvent = 0.5L . 1.1g/L = 0.55 g
We convert the mass to kg → 0.55 g . 1kg /1000g = 5.5×10⁻⁴ kg
m = mol/kg → 0.25 mol /5.5×10⁻⁴ kg = 454.5 m
If you are talking about just pure regular water, the answer is false. BUT, some salts dissolved IN WATER, can act as electrolytes. But regular water, no.
Answer:
- <u><em>Magnesium and fluorine.</em></u>
Explanation:
<em>Ionic compounds</em> are formed by the electrostatic attraction of cations and anions.
Cations, positive ions, are formed when atoms lose electrons, and anions, negative ions, are formed when atoms gain electrons.
When two different atoms have similar atraction for electrons (electronegativity) they will not donate to nor catch electrons from each other, so cations and anions will not be formed. Instead, the atoms would prefer to share electrons forming covalent bonds to complete their outermost shell (octet rule).
Then, in order to form ionic compounds the electronegativities have to substantially different. This situation does not happen between two nonmetal elements, which nitrogen and sulfur are. Then, you can predict safely that nitrogen and sulfur will not form an ionic compound.
Ionic compounds, then require the electronegativity difference that exist between some metals and nonmetals. Being magnesium an alkaline earth metal, its electronegativity is very low. On the other hand, fluorine the first element of the group 17, has the highest electronegativity of all the elements.Thus magnesium and fluorine will have enough electronegativity difference to justify the exchange of electrons, forming ions and, consequently, ionic compounds.
Answer:
[H2] = 0.0692 M
[I2] = 0.182 M
[HI] = 0.826 M
Explanation:
Step 1: Data given
Kc = 54.3 at 430 °C
Number of moles hydrogen = 0.714 moles
Number of moles iodine = 0.984 moles
Number of moles HI = 0.886 moles
Volume = 2.40 L
Step 2: The balanced equation
H2 + I2 → 2HI
Step 3: Calculate Q
If we know Q, we know in what direction the reaction will go
Q = [HI]² / [I2][H2]
Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]
Q =(n(HI)²) /(nH2 *nI2)
Q = 0.886²/(0.714*0.984)
Q =1.117
Q<Kc This means the reaction goes to the right (side of products)
Step 2: Calculate moles at equilibrium
For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI
Moles H2 = 0.714 - X
Moles I2 = 0.984 -X
Moles HI = 0.886 + 2X
Step 3: Define Kc
Kc = [HI]² / [I2][H2]
Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]
Kc =(n(HI)²) /(nH2 *nI2)
KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))
X = 0.548
Step 4: Calculate concentrations at the equilibrium
[H2] = (0.714-0.548) / 2.40 = 0.0692 M
[I2] = (0.984 - 0.548) / 2.40 = 0.182 M
[HI] = (0.886+2*0.548) /2.40 = 0.826 M
