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Goshia [24]
3 years ago
11

The alkali metals all have

Chemistry
1 answer:
qaws [65]3 years ago
6 0
D-1 electron in their outer energy level.
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Which of the following are strongly hydrogen bonded in the liquid phase? A) nitrilesB) esters C) secondary amides D) acid chlori
-Dominant- [34]

Answer: Option (C) is the correct answer.

Explanation:

Chemical formula of a secondary amide is R'-CONH-R, where R and R' can be same of different alkyl or aryl groups. Here, the hydrogen atom of amide is attached to more electronegative oxygen atom of the C=O group.

Therefore, the hydrogen atom will be more strongly held by the electronegative oxygen atom. As a result, there will be  strongly hydrogen bonded in the liquid phase of secondary amide.

Whereas chemical formula of nitriles is RCN, ester is RCOOR' and acid chlorides are RCOCl. As no hydrogen bonding occurs in any of these compounds because hydrogen atom is not being attached to an electronegative atom.

Thus, we can conclude that secondary amides are strongly hydrogen bonded in the liquid phase.

6 0
3 years ago
The isotope of an atom containing 40 proteins and 51 neutrons suddenly has two neutrons added to it what isotope is created?
aev [14]
The correct answer is (C. zirconium - 93.
7 0
3 years ago
The element in Group 14 Period 4 has how many neutrons and is it a metal, nonmetal, metalloid or noble gas.
monitta

Number of neutrons=41, and it is metalloid

<h3>Further explanation</h3>

Group 14⇒valence electron = 4(ns²np²)

Period 4⇒n=4

So electron configuration of the element :

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p² = 32 electron=atomic number

The element with atomic number 32, which is in period 4 and group 14 is Ge-Germanium

There are seven elements that can be classified as metalloids, namely boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po).

While the mass number: 73

So the number of neutrons = mass number-atomic number

\tt 73-32=41

8 0
2 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{20\text{ min}}

k=3.465\times 10^{-2}\text{ min}^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

a-x=0.100M

Therefore, the concentration of A after 80 min is, 0.100 M

3 0
2 years ago
Put the following steps in order to show how a scientist follows Scientific Method
Gekata [30.6K]
1.question, 2.observe, 3.hypothesize, 4.experiment 5.conclusion, 6. record.
8 0
2 years ago
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