The amount of heat lost by granite is equal to the amount
of heat gained by water. Therefore their change in enthalpies must be equal.
The opposite in sign means that one is gaining while the other is losing
ΔH granite = - ΔH water
ΔH is the change in enthalpy experienced by a closed object
as it undergoes change in energy. This is expressed mathematically as,
ΔH = m Cp (T2 – T1)
Given this information, we can say that:
12.5 g * 0.790 J / g ˚C * (T2 – 82 ˚C) =
- 25.0 g * 4.18 J / g ˚C
* (T2 – 22 ˚C)
9.875 (T2 – 82) = 104.5 (22 – T2)
9.875 T2 – 809.75 = 2299 – 104.5 T2
114.375 T2 = 3108.75
T2 = 27.18 ˚C
The temperature of 2 objects after reaching thermal
equilibrium is 27.18 ˚<span>C.</span>
Global warming, Cosmic Background radiation (even though most is blocked not ALL), and pollution.
Answer:
Pecan trees 1 and 4
Explanation:
1: 1 and 4
2a: With good soil and good parent plants
2b: Heredity
Answer:
Ok so, b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...
i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced
ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.
iii. Calculate the standard potential (voltage) of the cell
Look up the reduction potential,
E
⁰
red
, for the reduction half-reaction in a table of reduction potentials
Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,
E
⁰
ox
=
-
E
⁰
red
.
iv. What kind of electrochemical cell is this? Explain your answer.
All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells
1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery
Explanation:
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
