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alexira [117]
3 years ago
11

PLEASE HELP!!!

Chemistry
1 answer:
Inga [223]3 years ago
3 0

Answer:

d=1.5\ g/cm^3

Explanation:

Given that,

The mass of the rock, m = 15 g

The volume of the rock, V = 10 cm³

We need to find the density of the rock. The density of an object is equal to the mass per unit volume such that,

d=\dfrac{m}{V}\\\\d=\dfrac{15\ g}{10\ cm^3}\\\\d=1.5\ g/cm^3

So, the density of the rock is equal to 1.5\ g/cm^3.

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What characteristics separate the class Insecta from other classes of the phylum Arachnida
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Answer:

Three pairs of walking legs, wings, body divided into three segments, pair of sensory antennae.

Explanation:

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Using the model of the periodic table, which two elements pictured have similar chemical properties?
Murljashka [212]

Answer:

1 and 3

Explanation:

The vertical columns (groups) of the periodic table are arranged such that all its elements have the same number of valence electrons. All elements within a certain group thus share similar properties.

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The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
2 years ago
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The correct answer is (3)


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The explanation:


according to attached table:


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and They both have β- decay mode and with half-lives greater than hour.


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