Answer:
Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points.
In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.
Answer:
hey there! it gains an electron and has an octet in its inner shell it loses an electron and has an octet in the next lowest energy level it gains an electron and has an octet in the lowest energy level it gains an electron and has an octet in its outer shell
HAVE A GRET DAY! ;D
Explanation:
tbh idk if im 100% sure if im right but i hope i am :)
Answer:
Iron is a metal.
Iron wool is made up of thin strands of
iron loosely bundled together.
Your teacher has attached a piece of
iron wool to a see-saw balance. At the
other end of the see-saw is a piece of
Plasticine.
Iron wool can combust. Your teacher is
going to make the iron wool combust by
heating it.
If there is a change in mass, the see-saw
will either tip to the left or to the right.
Explanation:
Cu ions reduced to cu and Al oxidized to Al ions is the answer. copper ions is right hand side of the E.M.F cell where reduction take place and is reduced to copper metal. Al metal is at left hand side of the E.m.f cell where oxidation take place and is oxidized to Al ions
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
