A precipitate will be formed through the double displacement method in chemical reactions. The chemical equation would be:
Na2CO3 + AgNO3 ----> NaNO3 + Ag2CO3.
Balancing would result to:
<span>Na2CO3 + 2AgNO3 ----> 2NaNO3 (aq)+ Ag2CO3.
</span>
Ag2CO3 is the precipitate formed. It is yellow in color but due to the presence of elemental silver, it is a bit grayish.
Answer:
See explanation
Explanation:
We know that water begins to expand at 4°C. This is generally referred to as the anomalous behavior of water.
The implication of this is that ice has a greater volume than water. Thus ice is less dense than liquid water.
n-butyl alcohol has much less density than water. This accounts for the fact that water upon freezing breaks the bottle while n butyl alcholol leads to a container with concave walls.
Hence the density of ice is much higher than the density of n butyl alcohol.
Answer:
16.4 °C
Explanation:
Boiling point elevation is the phenomenon in which the boiling point of a solvent will increase when another compound is added to it; meaning that athe resultant solution has a higher boiling point than its pure solvent.
Using the ebullioscopic constant,
ΔT = m * i * Kb
Where,
Δ T is the temperature difference between the boiling point of the solution, Temp.f and boiling point of the pure solvent, Temp.i
Kb is the ebulliscope factor of water = 0.510 °C.kg/mol
i is the van hoffs number = 1
m is the molality in mol/kg.
Calculating the molality of the solution,
Temp.i = 100°C
Temp.f = 104.5 °C
= 4.5/(1*0.510)
= 8.8235 mol/kg
Freezing point depression is defined as the decrease in the freezing point of a solvent on the addition of a solute.
Using the same equation, but kf = 1.86 °C.kg/mol
ΔT = m * i * Kf
Temp.i = freezing point of water = 0°C
Temp.f = (8.8235*1.86) - 0
= 16.412 °C
Freezing point of the solution = 16.4 °C
Answer: pH of an 
solution is 4.34
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Putting in the values:
![pH=-\log[4.58\times 10^{-5}]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B4.58%5Ctimes%2010%5E%7B-5%7D%5D)
Thus pH of an solution if the ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
is ![4.58\times 10^{-5}]](https://tex.z-dn.net/?f=4.58%5Ctimes%2010%5E%7B-5%7D%5D)
is 4.34
