Answer:
0.5
Explanation:
1 mole of ammonium nitrate contains 2 moles of nirogen
1 mole of nitrogen converts to 0.5 moles of ammonium nitrate
the conversation factor is 0.5
Mass of iodine solid produced = 1.6 g
Iodine solid has a molecular formula : I2
Atomic mass of iodine = 126.904 g/mol
Therefore, molar mass of I2 solid = 2* 126.904 = 253.808 g/mol
Number of moles of a given substance is the ratio of its mass to the molar mass
# moles of I2 solid = mass of I2 solid/molar mass = 1.6 g/253.808 = 0.00630 moles
Hence, the # moles of iodine solid produced is 0.00630 moles
Answer:
The hydrogen spectrum is an important piece of evidence to show the quantized electronic structure of an atom. ... It results in the emission of electromagnetic radiation initiated by the energetically excited hydrogen atoms. The hydrogen emission spectrum comprises radiation of discrete frequencies.
The spectrum starts with red light, with a wavelength of 700 nanometers (7,000 angstroms), at the top. ... It spans the range of visible light colours, including orange and yellow and green, and ends at the bottom with blue and violet colours with a wavelength of 400 nm (4,000 angstroms).
Explanation:
Hydrogen molecules are first broken up into hydrogen atoms (hence the atomic hydrogen emission spectrum) and electrons are then promoted into higher energy levels. Suppose a particular electron is excited into the third energy level. It would tend to lose energy again by falling back down to a lower level.
The spectrum of the Sun appears as a continuous spectrum and is frequently represented as shown below. This type of spectrum is called an emission spectrum because what you are seeing is the direct radiation emitted by the source.
Light energy is turned into chemical energy when <span>when a photochemically excited special chlorophyll molecule of the photosynthetic reaction center loses an electron, undergoing an oxidation reaction.
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The generalized rate expression may be written as:
r = k[A]ᵃ[B]ᵇ
We may determine the order with respect to B by observing the change in rate when the concentration of B is changed. This can be done by comparing the first two runs of the experiment, where the concentration of A is constant but the concentration of B is doubled. Upon doubling the concentration of B, we see that the rate also doubles. Therefore, the order with respect to concentration of B is 1.
The same can be done to determine the concentration with respect to A. The rate increases 4 times between the second and third trial in which the concentration of B is constant, but that of A is doubled. We find that the order with respect to is 2. The rate expression is:
r = k[A]²[B]
