Answer:
1. filtration and evaporation
2. i) water is added to the sand and salt mixture
ii) then the mixture is filtrated and so the sand and the salt water was seperated
iii) the salt water is heated with the help of burner until the water gets evaporated
iv) after the water gets evaporated, the salt is remained in the container
3. observation:
- on adding water to the mixture, the salt is completely dissolved in the water
- when filtrated the sand is seperated from the salt water
- now the salt water when heated with the burner until the evaporation, the water is evaporated
- the salt is precipitated and remained in the container
4. cautions:
- while using the burner, we should be cautious with fire
- the container that is heated should be holded with the help of a cloth to avoid heat
Answer:
Volume of stock solution needed = 6.0299 mL
Explanation:
<u>
</u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.
This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.
<u>Data:</u>
M1 = 6.01 M stock solution concentration
M2 = 0.3624 M diluted solution concentration
V2 =100 mL diluted solution volume
V1 = ? stock solution volume
M1 * V1 = M2 * V2
Answer:
–36 KJ.
Explanation:
The equation for the reaction is given below:
2B + C —› D + E. ΔH = – 24 KJ
From the equation above,
1 mole of D required – 24 KJ of energy.
Now, we shall determine the energy change associated with 1.5 moles of D.
This can be obtained as illustrated below:
From the equation above,
1 mole of D required – 24 KJ of energy
Therefore,
1.5 moles of D will require = 1.5 × – 24 = –36 KJ.
Therefore, –36 KJ of energy is associated with 1.5 moles of D.
Answer: -
Lower the specific heat capacity of the metal, more the amount of heat would be required to raise the temperature to the same extent.
So for the same amount of heat added to 5.00 g samples of metals, the metal with the lowest specific heat capacity would experience the smallest temperature change.
For example, if the elements are Al, Au ,Cu and Fe, then Au would experience the smallest temperature change due to least specific heat capacity.
