Answer:
2 Mg + CO2 → 2 MgO
Explanation:
Magnesium ribbon burns in the presence of oxygen to produce Magnesium Oxide
Cells are too small to see with the naked eye.
It's pretty straight forward, use the cross-out method.
1) Microscopes MAGNIFY images, they don't color the cells. In fact, scientists have to use these chemicals to "stain" or color the cells to see them more easily through microscopes.
2) If the lenses of a microscope reduced the image of an organism to the size of a cell, you'd be seeing a very tiny human through your microscope, instead of actual cells.
3) Microscopes don't "trap" anything. In fact, scientists use plates or slides under microscopes to contain what they're studying.
Answer:
A.) Polymers are substances that have relatively large molecules
Answer:
Hg∧2+ has a negative standard entropy because the ions are highly solvated in aqueous phase; smaller the ionic size, the more highly it is surrounded by solvated ions. Therefore it will be in highly ordered state hence the entropy decreases.
Hg2 ^2+ has a positive standard entropy because the ionic size of Hg^2+ is smaller than Hg2 ^2+, so therefore the Hg^2+ is highly solvated and that means that it is in highly ordered state. Hg2 ^2+ is not highly solvated so it will have a positive entropy
Explanation:
The values of standard entropy of aqueous ions has a negative standard entropy because the ions are highly solvated in aqueous phase; smaller the ionic size, the more highly it is surrounded by solvated ions. Therefore it will be in highly ordered state hence the entropy decreases.
Hg2 ^2+ has a positive standard entropy because the ionic size of Hg^2+ is smaller than Hg2 ^2+, so therefore the Hg^2+ is highly solvated and that means that it is in highly ordered state. Hg2 ^2+ is not highly solvated so it will have a positive entropy
According to half life equation:
T(1/2) = ㏑2 / K1
when the T(1/2) = 4 min * 60 = 240 sec
by substitution:
240 = 0.6931 / K1
K1 = 2.9 x 10^-3
when the second T(1/2) = 20 sec, so to get K2:
T(1/2) = 0.6931 / K2
by substitution:
20 = 0.6931 / K2
∴K2 = 3.4 x 10^-2
so, we can get T2 by using this formula:
㏑ (K2/K1) = Ea/R (1/T1 - 1/T2)
by substitution:
㏑(3.4 x 10^-2)/(2.9 x 10^-3) = (24520 / 8.314) (1/298 - 1/T2)
∴ T2 = 396.7 K
= 396.7 - 273 = 123.7 °C
