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3 years ago

Based on table s, an atom of which element has the strongest attraction for electrons in a chemical bond?.

1 answer:

7 0

Based on Table S, the atom of element that has the strongest attraction for electrons in a chemical bond from the options is Chlorine.

<h3>What is Electronegativity?</h3>

Electronegativity can be regarded as the tendency that is been associated to an atom for attraction of electrons to itself.

However, Fluorine is considered as the most electronegative element but based on Table S, Chlorine has the strongest attraction for electrons in a chemical bond .

THESE ARE THE OPTIONS

1 aluminum

2 chlorine

3 magnesium

4 sulfur

Learn more about Electronegativity at;

brainly.com/question/21636614

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In quadrilateral $ABCD$, we have $AB=3,$ $BC=6,$ $CD=4,$ and $DA=4$. If the length of diagonal $AC$ is an integer, what are all

Natalka [10]

<h3>Answer:</h3>

AC ∈ {4, 5, 6, 7}

<h3>Explanation:</h3>

When two sides of a triangle are specified, the allowable values for the length of the third side are between the sum and difference of the given sides.

In ∆ACD, sides DA and CD are both given as length 4. Thus the possible range of values for side AC is 0 – 8. The extremes of this range result in ∆ACD having zero area, so we assume they are not of interest.

In ∆ACB, sides AB and BC are given as having lengths 3 and 6. Thus the possible range of values for side AC is 3 – 9. The extremes of this range result in ∆ACB having zero area, so we assume they are not of interest.

The integers that are in both ranges and that give triangles with non-zero area are ...

... AC ∈ {4, 5, 6, 7}

5 0

3 years ago

Read 2 more answers

When the chemical reaction 2no(g)+o2(g)→2no2(g) is carried out under certain conditions, the rate of disappearance of no(g) is 5

viktelen [127]

The rate of disappearance of O2(g) under the same conditions is 2.5 × 10⁻⁵ m s⁻¹.

<h3>What is the rate law of a chemical equation? </h3>

The rate law of a chemical reaction equation is usually dependent on the concentration of the reactant species in the equation.

The chemical reaction given is;

$\mathbf{2 NO_{(g)} + O_{2(g)} \to 2 NO_{2(g)} }$

The rate law for this reaction can be expressed as:

$\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}= +\dfrac{1}{2}\dfrac{d[NO_2]}{dt}}$

Recall that:

The rate of disappearance of NO(g) = 5.0× 10⁻⁵ m s⁻¹.

Since both NO and O2 are the reacting species;

Then:

The rate of disappearance of NO(g) is equal to the rate of disappearance of O2(g)

$\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}}$

$\mathbf{= -\dfrac{1}{2} \times 5.0 \times 10^{-5} = rate \ of \ disappearance \ of \ O_2}$

Thus;

The rate of disappearance of O2 = 2.5 × 10⁻⁵ m s⁻¹.

Therefore, we can conclude that two molecules of NO are consumed per one molecule of O2.

Learn more about the rate law here:

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