The redox reaction is
Here
Calcium undergoes reduction, and acts as cathode
Lithium undergoes oxidation and acts as anode
The reduction potential of calcium is -2.87 V
The reduction potential of lithium is - -3.05 V
We know that
Ecell = Ecathode - Eanode
Ecell = -2.87 - (-3.05) = 0.18 V
<span>Answer:
For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees.
4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ.
Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work.
To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3.
.0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
Answer:
with the help of the juice contained in it
Answer:
2.25×10¯³ mm.
Explanation:
From the question given above, we obtained the following information:
Diameter in micrometer = 2.25 μm
Diameter in millimetre (mm) =?
Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m
2.25 μm = 2.25×10¯⁶ m
Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:
1 m = 1000 mm
Therefore,
2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm
2.25×10¯⁶ m = 2.25×10¯³ mm
Therefore, 2.25 μm is equivalent to 2.25×10¯³ mm.
Answer: 8.1 days
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant = x
a - x = amount left after decay process= 
a) to find rate constant
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
b) for completion of one fourth of reaction
Thus after 8.1 days , one fourth of original amount will remain.
