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2 years ago

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Question 2 (Mandatory) (2 points)

1 answer:

Deffense [45]2 years ago

4 0

Electrons have a positive charge is the right answer

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Chlorophenols impart unpleasant taste and odor to drinking water at concentrations as low as 5 mg/m3. They are formed when the c

trapecia [35]

Answer:

See explaination

Explanation:

Chlorine’s disinfection properties have helped improve the lives of billions of people around the world. Chlorine also is an essential chemical building block, used to make many products that contribute to public health and safety, advanced technology, nutrition, security and transportation.

Please kindly check attachment for the step by step solution of the given problem.

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2 years ago

Which is one way that topographic maps are different from other maps?

anzhelika [568]

One way they are different is that they show elevation

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2 years ago

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DIA [1.3K]

V_1=49.8mL
T_1=18°C=291K
T_2=83°C=356K

Using Charles law

$\\ \sf\longmapsto V_1T_2=V_2T_1$

$\\ \sf\longmapsto V_2=V_1T_2\div T_1$

$\\ \sf\longmapsto V_2=\dfrac{49.8(356)}{291}$

$\\ \sf\longmapsto V_2=\dfrac{17728.8}{291}$

$\\ \sf\longmapsto V_2=60.9mL$

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2 years ago

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

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

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

$\mathbf{E_3 = 16.041 \ meV}$

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2 years ago

Determine a massa de um átomo de carbono12 em gramas

lapo4ka [179]

I believe it would be 1.660539040 × 10−24 gram.

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2 years ago

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