<span>We need to calculate the equivalent amount in units of moles of ammonium ions from the mass units. For this we need the molar mass of the substances involved. We calculate as follows:
31.3 g </span>(NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol <span>ammonium ions</span>
A conversion factor is a fraction or a ratio representing a relationship of two different measurement values. To write 20% m/v to a conversion factor, we need to remember that a percent is a value that represents the amount of a part per 100 units of the whole. M/v in the given value represents that the percentage is by mass per volume. So, to write it as a conversion factor, we do as follows:
20% m/v = 20 mass units / 100 volume units = 1 mass units / 5 volume units
Usually units of this are in g per L. So, it is equivalent to 1 g / 5 L
<span>Waves transfer energy through
vibration. just like electromagnetic waves, energy is transferred through
vibrations of electric and magnetic fields. In sound waves, energy is
transferred through vibration of air particles or particles of a solid through
which the sound travels. In water waves, energy is transferred through the
vibration of the water particles. While particles transfer energy through
conduction and convection.</span>
Answer:
if chemistry hadn't been put up to practical use, we wouldn't truly understand the reason why humans are humans like makes up humans (including other understandings of biology, etc) and we wouldn't be able to have the advancements we have today (vaccines, etc).
Explanation:
<u>Answer:</u> The correct answer is 
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.
Chromium will undergo oxidation reaction and will get oxidized.
The half reactions for the above cell is:
Oxidation half reaction: 
Reduction half reaction: 
( × 3)
Net equation: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
Hence, the correct answer is
