When the amount of heat gained = the amount of heat loss
so, M*C*ΔTloses = M*C* ΔT gained
when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.
∴( M * C * ΔT )W = (M*C*ΔT) Al
when Mw is the mass of water = 100 g
and C the specific heat capacity of water = 4.18
and ΔT the change in temperature for water= 28-25 = 3 ° C
and ΔT the change in temperature for Al = 100-28= 72°C
and M Al is the mass of Al block
C is the specific heat capacity of the block = 0.9
so by substitution:
100 g * 4.18*3 = M Al * 0.9*72
∴ the mass of Al block is = 100 g *4.18 / 0.9*72
= 19.35 g
<h3><u>Answer</u>;</h3>
= 4.68 K
<h3><u>Explanation</u>;</h3>
According to the combined gas law;
P1V1/T1 = P2V2/T2
Given; P1 = 125 Psi
V1 = 75 L
T1 = 288 K
P2 = 25 PSI
V2 =6.1 L
Therefore;
T2 = P2V2T1/P1V1
= (25×6.1 ×288)/(125×75)
= 4.6848
= 4.68 K
Answer:
to which cations from the salt bridge migrate
Explanation:
A voltaic cell is an electrochemical cell that uses spontaneous redox reactions to generate electricity. It's composed of a cathode, an anode, and a salt bridge.
In cathode, the substance is gaining electrons, so it's reducing, in the anode, the substance is losing electrons, so it's oxidating. The flow of electrons is from the anode to the cathode.
The salt bridge is a bond between the cathode and the anode. When the redox reaction takes place, the substances produce its ions, so the solution is no more neutral. The salt bridge allows the solutions to become neutral and the redox reaction continues.
So, the cathode produces anions, which goes to the anode, and the anode produces cations, which goes to the cathode. Then, the cathode n a voltaic cell is the electrode to which cations from salt bridge migrate and where the reduction takes place.
Answer: the HCO3- to act as a base and remove excess H by the formation of H2CO3
Explanation:
H2CO3 in an aqueous solution is a buffer. This means the reaction is the following:
H2CO3 ------ HCO3- + H+
Then, the HCO3- that was formed acts as a base (absorbing a proton) like this
HCO3- + H+ ------- H2CO3
If there was an increase in H+, there would be an increase in the second reaction in an effort to neutralize that acid, thus making the H2CO3 more concentrated
Answer: 
Explanation:
Here Mn undergoes oxidation by loss of electrons, thus act as anode. silver undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials.
![E^0_{[Mn^{2+}/Mn]}= -1.18V](https://tex.z-dn.net/?f=E%5E0_%7B%5BMn%5E%7B2%2B%7D%2FMn%5D%7D%3D%20-1.18V)
![E^0_{[Ag^{2+}/Ag]}=+0.80V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B2%2B%7D%2FAg%5D%7D%3D%2B0.80V)
![E^0=E^0_{[Ag^{+}/Ag]}- E^0_{[Mn^{2+}/Mn]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-%20E%5E0_%7B%5BMn%5E%7B2%2B%7D%2FMn%5D%7D)
The standard emf of a cell is related to Gibbs free energy by following relation:
= gibbs free energy
n= no of electrons gained or lost = 2
F= faraday's constant
= standard emf = 1.98V
Thus the value of is
