Answer:
81.04°C
Explanation:
Heat loss by water = Heat gained by Aluminum
Heat loss by water;
H = MCΔT
ΔT = 100 - T2
M = 580g
c = 4.2
H = 580 * 4.2 (100 - T2)
H = 243600 - 2436T2
Heat ganed by Aluminium
H = MCΔT
ΔT = T2 - 24
M = 900g
c = 0.9
H = 900 * 0.9 (T2 - 24)
H = 810 T2 - 19440
243600 - 2436T2 = 810 T2 - 19440
243600 + 19440 = 810 T2 + 2436T2
263040 = 3246 T2
T2 = 81.04°C
Assumption;
Assume that energy diffuses throughout the pan and water so that all parts reach the same final temperature.
Are produced 72 grams of water in this reaction.
<h3>Mole calculation</h3>
To find the value of moles of a product from the number of moles of a reactant, it is necessary to observe the stoichiometric ratio between them:
Analyzing the reaction, it is possible to see that the stoichiometric ratio is 1:2, so we can perform the following expression:
So, if there are 2 mols of Ca(OH)2:
Ca(OH)2 | H2O
Finally, just find the number of grams of water using your molar mass:
So, 72 grams are produced of water in this reaction.
Learn more about mole calculation in: brainly.com/question/2845237
<span>The only scenario that
will allow you to reach an equilibrium mixture involving these chemicals is to
place NH3 into a sealed vessel. This reaction requires pressures between 2100,
3600 psi, and temperatures between 300 and 550 degree Celsius. With this given
temperature and pressure, the ammonia naturally decomposes into nitrogen and
hydrogen gas at the same rate. When this happen, the concentrations of these
chemicals become constant and the system is said to be at equilibrium.</span>
In order to find the NET nuclear charge from an atom's valence electron to the proton nucleus, you need to do some simple math in order to find the charge.
You would also need to apply math when you need to figure out what the charge on an ion is by either adding or taking away electrons depending on whether it is an anion or cation.
Answer:
The correct answer is Option C (E1) and Option B (carbocation).
Explanation:
- Intramolecular immunity idols are considered as that of the formation mechanism with E1 responses or reactivity.
- Reactants with E1 were indeed obligations of both parties, meaning that an E1 reaction was conducted thru all the two stages known as ionization but rather deprotonation. Involves the absence of either an aromatic ring, a carbocation has been generated throughout the ionization solution.
Some other possibilities offered aren't relevant to the procedure outlined. So the above alternative is accurate.
