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1 year ago

12

[H+]=1×10−13 M pH= what is this answeer

1 answer:

pogonyaev1 year ago

4 0

Answer:

Take the -logarithm of the concentration of hydrogen ions that are in the solution:

pH=−log(1×10−8M)=8.❤

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The carbohydrates seen here contains 3 common elements.They are....

Llana [10]

The answer is A),<span>carbon,hydrogen,and oxygen

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2 years ago

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Which describes the state at which products form at the same rate as reactants?

swat32

Chemical equilibrium<span> is the state in which both reactants and products are present in concentrations which have no further tendency to change with time.
</span><span>Or, we can say that in chemical equilibrium the ratio between the concentration of the reactants and the products is constant.</span><span>
Chemical equilibrium is a result state when </span><span>the forward reaction proceeds at the same rate as the reverse reaction.
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2 years ago

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Density of water is 1 g/ml. If a piece of wood with a density of 5 g/ml is

Alchen [17]

Answer:

The piece of wood will sink

Explanation:

If an object has a density higher than water, it will sink, and the piece of wood is more dense than water when measured.

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

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

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

$\mathbf{E_3 = 16.041 \ meV}$

8 0

2 years ago

Exactly how many -OH groups are found in a 6 carbon alditol?

mr Goodwill [35]

Answer:

The number of OH group found in found in a 6 carbon alditol is 6

Explanation:

The structure of a 6 carbon alditol is shown on the first uploaded image

8 0

2 years ago