Answer:
1. 2.510kJ
2. Q = 1.5 kJ
Explanation:
Hello there!
In this case, according to the given information for this calorimetry problem, we can proceed as follows:
1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:
Then, we perform the conversion as follows:
2. Here, we use the general heat equation:
And we plug in the given mass, specific heat and initial and final temperature to obtain:
Regards!
For the answer to the question above, asking to w<span>rite the complete balanced equation for the reaction between aluminum metal (Al) and oxygen gas (O2)and You do not need to make the subscripts smaller.
My answer would be,
</span><span>4Al(s) + 3O2(g) --->2 Al2O3(s)
</span>
I hope this helps.
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Well you didn’t post any but I’m hoping that either plastic, lead, wood, glass, or paper, are an option.
Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium
Heat gained by water = 100 × 4.186 × 30.5
= 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5
= 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules
