Follow
these steps to solve the given equation:
Multiply
the two decimal figures together and find the sum of the exponents, that is,
(1.5
* 1.89) * 10 ^4+3
(2.835)
* 10^7
10^7
can also be written as e.70
'e'
stands for exponential.
Therefore,
we have 2. 835 e 7.0 = 2.8 e 7.0.
Based on the calculations above, the correct option is A.
The formula for energy or enthalpy is:
E = m Cp (T2 – T1)
where E is energy = 63 J, m is mass = 8 g, Cp is the
specific heat, T is temperature
63 J = 8 g * Cp * (340 K – 314 K)
<span>Cp = 0.3 J / g K</span>
The first question's answer is
Explanation:
As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.
Hence, amount of diethyl ether present will be calculated as follows.
(100ml - 1.468 ml)
= 98.532 ml
So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.
Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.
Amount of water = 
= 0.3427 ml
Now, when magnesium dissolves in water then the reaction will be as follows.

Molar mass of Mg = 24.305 g
Molar mass of
= 18 g
Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.
Amount of Mg =
= 0.462 g
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V