Mass SrBr2 = 247.42 g/mol x 1.36 mol
<span>= 336 g</span>
Answer: product
Explanation:
Each substance written to the right of the arrow in a chemical equation is referred to as a product.
When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.
On the other hand which is the right side is the product.
Answer:
Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)
Explanation:
If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.
The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.
On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.
Answer:
Various limitations of Mendeleev's periodic table are:-
Position of hydrogen - he couldn't assign a correct position to hydrogen as it showed properties of both alkali and halogens .
Position of isotopes - he considered that the properties of elements are a function of their atomic masses. Hence isotopes of a same element couldn't be placed.
In the d-block , elements with lower atomic number were placed before higher atomic number.
Explanation:
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or
)
Given : kf = 1.86 
m = 0.907
)
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86
* 0.907
)
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C