Answer:
The answer is a carbon isotope.
Explanation:
Carbon has 6 protons and electrons so the atom is neutral.
However it is a carbon isotope because it has more neutrons than protons so be aware of that
First, we need to find the atomic mass of

.
According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16
As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The
molar mass of

= 6 * 12.01 + 12 * 1.008 + 6 * 16
The
molar mass of

= 180.156
grams/moleNow that we have the molar mass of

, we can find the grams of glucose by using:
mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)
Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams
Ans: Mass of glucose in grams in 2.47 moles =
444.99 grams
-i
Where are the following answers?
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
<em>Alkali metals are among the most reactive metals. This is due in <u>part to their larger atomic radii and low ionization energies.</u> They tend to donate their electrons in reactions and have an oxidation state of +1. ... All these characteristics can be attributed to these elements' large atomic radii and weak metallic bonding.</em>
Explanation:
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