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Naya [18.7K]
3 years ago
10

Can an ether form hydrogen bonds in the same way as does an ester

Chemistry
1 answer:
belka [17]3 years ago
5 0

Yes, an OH group from ethanol can form a hydrogen bond to the ether O atom in the same way as it can do so with the single-bonded O atom in the ester.

The O atom in the carbonyl group of the ester can also form H-bonds with ethanol.


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Natali5045456 [20]

Answer:

The last answer is right they get half of the mothers genes and half of the fathers genes.

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3 years ago
What do cations and anions form?
melomori [17]

Answer: Cations (positively-charged ions) and anions (negatively-charged ions) are formed when a metal loses electrons, and a nonmetal gains those electrons. The electrostatic attraction between the positives and negatives brings the particles together and creates an ionic compound, such as sodium chloride.

7 0
3 years ago
Given the unbalanced equation below, answer the following: Calculate the number of liters of 3.00 M lead (II) iodide solution pr
mr_godi [17]

The number of liters of 3.00 M lead (II) iodide : 0.277 L

<h3>Further explanation</h3>

Reaction(balanced)

Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)

moles of KI = 1.66

From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :

\tt \dfrac{1}{2}\times 1.66=0.83

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large \boxed {\bold {M ~ = ~ \dfrac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):

\tt V=\dfrac{n}{M}\\\\V=\dfrac{0.83}{3}\\\\V=0.277~L

5 0
3 years ago
Write the balanced equation showing the decomposition of carbonic acid and sulfurous acid.
Travka [436]

Explanation:

here's the answer to your question

4 0
3 years ago
500ml of a buffer solution contains 0.050 mol nahso3 and 0.031
nydimaria [60]

Answer:

The answers are explained below

Explanation:

a)

Given: concentration of salt/base = 0.031

concentration of acid = 0.050

we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59

b)

we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O

Moles i............0.05...................0.01.................0.031.....................0

Moles r...........-0.01.................-0.01................0.01........................0.01

moles f...........0.04....................0....................0.041.....................0.01

c)

we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041

Hence, we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71

d)

pOH = -log (0.01/0.510) = 1.71

pH = 14 - 1.71 = 12.29

e)

Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.

4 0
4 years ago
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