Colorimetric methods of analysis make use of Color changes in reagents to decipher the concentration of solutions.
Therefore, color completes the sentence
<h3>Colourimetric analysis</h3>
Generally, Colorimetric analysis is used to know the concentration of a chemical element in a solution while using colour indicators or reagents.
Therefore
Colorimetric methods of analysis are based on having a reagent that changes Color as a function of the concentration of the analyte.
Color
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Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:
A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.
Therefore the only choice that fits is A. 80 g of Br2.
Stirring this is because the three elements are factors affecting dissolving of a solvent. Eg temprature affects in hotness or coldness, Particle size affects whether it is big or small while quantity of soluble affects by the amount
You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!
group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1
we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!
Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.
Answer:
12 atm
Explanation:
First, let us convert Celcius into Kelvin: 28.0 °C = 301.15 K and 129.0 °C = 402.15 K
For this question we must employ the Combined Gas Law: 
, where 
is the initial pressure and 
is the new pressure.
We know that intitially, P=9 atm, V=30 L, and T=301.15K. From our problem, only temperature and pressure changes, while the number of moles, volume and the gas constant, R, stay the same, so they are irrelevant.
Thus, the filled out Combined Gas Law would be:
=
, where the volume, moles of gas, and R are cancelled out.
We can manipulate this equation to derive the new pressure. We find that
9atm≈0.74885.
This means that
≈9/0.74885≈12 atm
