Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.
Answer:

Explanation:
The formula for molality is:

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

Substitute the values into the formula.

Divide.

The molality is <u>2.8 moles per kilogram</u>
Answer:
The significance of "Er" in the diagram is :
B.) Threshold energy for reaction
Explanation:
Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .
Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.
<u>Activation Energy = Threshold Energy - Average Kinetic Energy</u>
<u>This means Activation energy decreases on increasing kinetic energy</u>
On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.
Catalyst also reduces the Activation energy.
<u>Er = Threshshold energy for reaction at 30 degree</u>
<u>Ea = Activation Energy</u>
<u>The given figure shows that the threshold energy decreases on increasing the temperature</u>
<u>Only the molecule having energy greater than Er can react to form product</u>
Answer:
5.0 38 84.0 749.7 528.0 729.0 738.9 739.0
Answer:
3.07 Cal/g
Explanation:
Step 1: Calculate the heat absorbed by the calorimeter
We will use the following expression.
Q = C × ΔT
where,
- C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
- ΔT: temperature change (2.29 °C)
Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ
According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.
Step 2: Convert 86.1 kJ to Cal
We will use the conversion factor 1 Cal = 4.186 kJ.
86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal
Step 3: Calculate the number of Cal per gram of candy
20.6 Cal/6.70 g = 3.07 Cal/g