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3 years ago

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Element R and Element Q have the same number of valence electrons. These elements have similar chemical behavior, but element R

has fewer energy levels than element Q
Describe what their positions on the Periodic Table have in common.

1 answer:

I am Lyosha [343]3 years ago

7 0

Answer:

I dont now because I am in 6 class

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Calculate the pH of a 0.50 M HIO. The Ka of hypoiodic acid, HIO, is 2.3x10–11.0.305.325.479.474.80

never [62]

Answer:

pH = 5.47

Explanation:

The equilibrium that takes place is:

HIO ↔ H⁺ + IO⁻

Ka = $\frac{[H+][IO-]}{[HIO]}$ = 2.3 * 10⁻¹¹

At equilibrium:

[HIO] = 0.5 M - x

[H⁺] = x

[IO⁻] = x

<u>Replacing those values in the equation for Ka and solving for x:</u>

$Ka=\frac{x^2}{0.5-x}=2.3*10^{-11} \\x^2=(2.3*10^{-11})(0.5-x)\\x^2=1.15*10^{-11}-2.3*10^{-11}x\\x^2+2.3*10^{-11}x-1.15*10^{-11}=0\\x=3.39*10^{-6}$

Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47

7 0

3 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

7 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

Practice Problem 07.61 Draw the major product that is obtained when (2S,3S)-2-Bromo-3-phenylbutane is treated with sodium ethoxi

schepotkina [342]

Answer:

Look at the picture.

Explanation:

(2S,3S)-2-Bromo-3-phenylbutane will undergo E2 reaction and form trans product of elimination due to its thermodynamic stability.

7 0

3 years ago

Need helpppp please

Vladimir79 [104]

Answer:

message me so i can help you because i cant see the words

Explanation:

6 0

3 years ago