It is unique because we can drink from it .
Answer:
Single replacement reaction (aka single displacement reaction)
Explanation:
In a single replacement reaction, one element is substituted for another in a compound to create a new compound and a new element in the products. The general form is:
A + BC --> B + AC
In the case of this question, Cr and Fe "trade places."
Answer:
Q = 0.061 = Kc
Explanation:
Step 1: Data given
Temperature = 500 °C
Kc=0.061
1.14 mol/L N2
5.52 mol/L H2
3.42 mol/L NH3
Step 2: Calculate Q
Q=[products]/[reactants]=[NH3]²/ [N2][H2]³
If Qc=Kc then the reaction is at equilibrium.
If Qc<Kc then the reaction will shift right to reach equilibrium.
If Qc>Kc then the reaction will shift left to reach equilibrium.
Q = (3.42)² / (1.14 * 5.52³)
Q = 11.6964/191.744
Q = 0.061
Q = Kc the reaction is at equilibrium.
Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
Answer:
The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>
Explanation:
The formula for molal boiling Point elevation is :
= elevation in boiling Point
= Boiling point constant( ebullioscopic constant)
m = molality of the solution
<em>i =</em> Van't Hoff Factor
Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .
In solution Mg3(PO4)2 dissociates as follow :
Total ions after dissociation in solution :
= 3 ions of Mg + 2 ions of phosphate
Total ions = 5
<em>i =</em> Van't Hoff Factor = 5
m = 8.5 m
= 0.512 °C/m
Insert the values and calculate temperature change:
Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K
= 373.15 K[/tex]
21.76 = T - 373.15
T = 373.15 + 21.76
T =394.91 K
