Depending what you have available there are a few different chemicals you can combine to show an obvious reaction without special equipment. Vinegar and Baking soda is a common mixture. Perhaps something like HCl and NaOH mixed will produce heat.
Answer:
a) The mass of lead in the alloy = 0.28 kg
b) There can be dissolved 1.46 kg more of lead.
Explanation:
Step 1: Data given
mass of the magnesium-lead alloy = 5.5 kg
200 °C → 5 wt% Pb
350 °C → 25 wt% Pb
a) What mass of lead is in the alloy?
We have to calculate (for the magnesium-lead alloy) the mass of lead in 5.5 kg of the solid α phase at 200°C just below the solubility limit. The solubility limit for the α phase at 200°C is about 5 wt% Pb.
The mass of lead in the alloy = (0.05)*(5.5 kg) = 0.28 kg
b) If the alloy is heated to 350°C, how much more lead may be dissolved in the α phase without exceeding the solubility limit of this phase?
At 350°C, the solubility limit of the a phase increases to approximately 25 wt% Pb.
C(lead) = ((mass fo lead in alloy) + (mass lead)) / ((magnesium-lead alloy mass) + (mass lead))
0.25 = (0.28 + m(Pb)) / ( 5.5 + m(Pb))
0.25 * ( 5.5 + m(Pb)) = (0.28 + m(Pb))
1.375 + 0.25 m(Pb) = 0.28 + m(Pb)
1.095 = 0.75 m(Pb)
m(Pb) = 1.095 / 0.75
m(Pb) = 1.46
There can be dissolved 1.46 kg more of lead.
Answer:
In the Bohr model, electrons can exist only in certain energy levels surrounding the atom.
When electrons jump from a higher energy level to a lower one, they emit light at a wavelength that corresponds to the energy difference between the levels.
The energy levels in each atom are unique.
Explanation:
Be sure to put in your own words.
Edge 2020
~theLocoCoco
W*V i believe because it comes down to this :
I is the current, W is wattage V is volts and the * is the thing that represents the Amperes (i’m not 100% but this is my best)
Use the formula E=hv, h=plancks constant and v=frequency
use the formula c=v*lambda to find v
the answer will be 2.88*10^-23J