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brilliants [131]
2 years ago
11

How many atoms is 3.49x1032 moles of KOH?

Chemistry
2 answers:
antoniya [11.8K]2 years ago
7 0
2.10098*10^47 atoms
Because no. Of atoms = no. Of moles * avogadros no

8090 [49]2 years ago
4 0

2.1 x 10⁵⁶ atoms

<h3>Further explanation </h3>

The mole is the number of particles(molecules,atoms,ions) contained in a substance

1 mol = 6.02.10²³

number of atoms :

\tt 3.49\times 10^{32}\times 6.02\times 10^{23}=2.1\times 10^{56}

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Write the equations that represent the first and second ionization steps for hydroselenic acid (H2Se) in water. (Use H3O+ instea
Otrada [13]

Answer:

The equations are

1) H_{2}Se+H_{2}O--> H_{3}O^{+}+HSe^{-}

2) HSe^{-} +H_{2}O--> H_{3}O^{+}+Se^{-2}

Explanation:

There are two ionization steps in the dissociation of hydroselenic acid.

In first dissociation the H₂Se loses one proton and forms hydrogen selenide ion as shown below:

H_{2}Se+H_{2}O--> H_{3}O^{+}+HSe^{-}

The next step is again removal of a proton from the base formed above.

HSe^{-} +H_{2}O--> H_{3}O^{+}+Se^{-2}

4 0
2 years ago
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answer:

no

explanation:

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Which electron has a higher probability of being found at 4 Å of the nucleus, one in a 2s orbital or one in a 2p orbital?
Lena [83]

Answer:

one in a 2s orbital

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Because of the peak near the nucleus in the 2s curve there is a higher probability of finding a 2s within 4 Å of the nucleus. In a multi-electron atom an electron in a 2s orbital will have a lower energy than one in a 2p orbital

4 0
2 years ago
Identify one disadvantage to each of the following models of electron configuration:
Murrr4er [49]

Answer:

The disadvantages of each of the given model of electron configuration have been mentioned below:

1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.

2). Arrow and line diagrams make the counting of electrons and take up too much space.

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6 0
2 years ago
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vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abu
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Explanation:

Average atomic mass of the  vanadium = 50.9415 amu

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Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

The atomic mass of isotope (II) of vanadium is 49.944 amu.

5 0
2 years ago
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