Answer:
The final temperature will be close to 20°C
Explanation:
First of all, the resulting temperature of the mix can't be higher than the hot substance's (80°C) or lower than the cold one's (20°C). So options d) and e) are imposible.
Now, due to the high heat capacity of water (4,1813 J/mol*K) it can absorb a huge amount of heat without having a great increment in its temperature. On the other hand, copper have a small heat capacity (0,385 J/mol*K)in comparison.
In conclusion, the copper will release its heat decreasing importantly its temperature and the water will absorb that heat resulting in a small increment of temperature. So the final temperature will be close to 20°C
<u>This analysis can be done because we have equal masses of both substances. </u>
3+
So, compounds of boron contain boron in a positive oxidation state, generally +3. The sum of oxidation numbers of all constituent atoms of a given molecule or ion is equal to zero or the charge of the ion, respectively. ... In most of the stable compounds of boron, its oxidation number is +3
The particles of objects have
both kinetic and potential energy because these forces are drive by the force
of motion or stillness of an object. Potential energy is the a type of energy
which an object possess however without motion. Kinetic energy in the other
hand, is the energy in motion or if the object moves along from one space to
another with respect to time. They both have these two energies by the presence
of atoms in these entities.
This doesn't need an ICE chart. Both will fully dissociate in water.
Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.
Step 1:
write out balanced equation for the reaction
HClO4+KOH ⇔ KClO4 + H2O
the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4
Step 2:
Determining the number of moles present in HClO4 and KOH
Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4
Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L
Remember:
M = moles/L so we have 0.025 L of 0.723 moles/L HClO4
Multiply the volume in L by the molar concentration to get:
0.025L x 0.723mol/L = 0.0181 moles HClO4.
Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH
Step 3:
Determine how much HClO4 remains after reacting with the KOH.
Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:
moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0
This means all of the HClO4 is used up in the reaction.
If all of the acid is fully reacted with the base, the pH will be neutral = 7.
Determine the H3O+ concentration:
pH = -log[H3O+]; [H3O+] = 10-pH = 10-7
The correct answer is 1.0x10-7.
Answer: I go search some information
Explanation:I come back with a answer
