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SIZIF [17.4K]
3 years ago
13

QUESTION 8

Chemistry
1 answer:
Alona [7]3 years ago
8 0
1. combustion
2.decompostioni
3.synthesids
4double replacement
5.single replacement
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using the law of conservation of energy explain three types of energy transformations that occur during a roller coaster ride fr
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According to valence bond theory, which orbitals overlap in the formation of the bond in hcl?
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According  to  valence  bond  theory    sigma  bonds   is  formed when   two  orbitals  approach  and overlap  over  each  other   while     pie  bonds  is  formed  when   two  orbitals  overlap  side  by  side. in formation  of    HCl   1s  orbital  of  hydrogen  overlap  on   3p  orbitals  of  chlorine
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A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

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.

           

3 0
3 years ago
What property of water allows it to stick to a dry surface
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You are looking for ADHESION
4 0
4 years ago
How many liters of 1.75 M solution could be made using 35 grams of NaCl?
dolphi86 [110]
Data:
M (molarity) = 1.75 M (mol/L)
m (mass) = 35 g
MM (molar Mass) of NaCl = 58.44 g/mol
V (volume) = ? (in liters)

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
1.75 =  \frac{35}{58.44*V}
1.75*58.44V = 35
102.27V = 35
V =  \frac{35}{102.27}
\boxed{\boxed{V \approx 0.34\:L}}\end{array}}\qquad\quad\checkmark
3 0
3 years ago
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