Answer:
There are 3, 64 moles of NaCl.
Explanation:
First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate themoles in 213 grams of NaCl, making a simple rule of three:
Weight NaCl= Weight Na + Weight Cl = 23 g + 35, 5 g= 58, 5 g/ mol
58,5 g ------1 mol NaCl
213 g---------x= (213 g x 1 mol NaCl)/ 58, 5 g= <em>3, 64 mol NaCl</em>
Answer:
B
Explanation:
Ionic compound can conduct electricity
Answer:
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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: electrons
Explanation: Electrons have a charge of -1 each. If two left, the remaining atom would have a positive +2 change.
