Answer:
hm step by step try to figure it 9ut
The reaction → is best classified as double displacement reaction.
Those reaction in which two compounds react by exchanges of ions to form two new compounds is called double displacement reaction. The easiest way to identify double displacement reactions is to check to see whether the cations exchanged anions with each other or not . Always balanced chemical equation is used to determine.
There are three types of double displacement reaction which is given as,
- Precipitation
- Neutralization
- Gas formation
The real world example of double displacement reaction is combining vinegar and baking soda to create homemade volcano.
learn more about double displacement reaction
brainly.com/question/13870042?
#SPJ4
Answer is: boiling point will be changed by 4°C.
Chemical dissociation of aluminium nitrate in water: Al(NO₃)₃ → Al³⁺(aq) + 3NO⁻(aq).
Change in boiling point: ΔT =i · Kb · b.
Kb - molal boiling point elevation constant of water is 0.512°C/m, this the same for both solution.
b - molality, moles of solute per kilogram of solvent., this is also same for both solution, because ther is same amount of substance.
i - Van't Hoff factor.
Van't Hoff factor for sugar solution is 1, because sugar do not dissociate on ions.
Van't Hoff factor for aluminium nitrate solution is approximately 4, because it dissociates on four ions (one aluminium cation and three nitrate anions). So ΔT is four times bigger.
Answer:
q1..no.2 and 4 are aromatic
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C