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Tanya [424]
3 years ago
11

How does the number of valence electrons for elements change across a period?

Chemistry
2 answers:
kondor19780726 [428]3 years ago
8 0

Answer: Option (b) is the correct answer.

Explanation:

Valence electrons is the number of electrons present in the outermost shell of an atom.

So, when we move across a period from left to right then there will be increase in number of valence electrons.

For example, valence electrons in carbon is 4, nitrogen has 5 valence electrons, oxygen has 6 valence electrons.

Thus, we can conclude that number of valence electrons increases for elements change across a period.

NeTakaya3 years ago
4 0
D. The number increases and then decreases for noble gases
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A. helium, neon and argon, because they are in the same group or column
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fill in the blank: 1.)the amount of space that matter fills is its.................. 2.)a state of matter with a definite volume
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1) mass

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The survival of aquatic organisms depends on the small amount of O2 that dissolves in H2O . The diagrams above represent possibl
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Answer:

Diagram 1

Explanation:

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Water is a polar molecule having oxygen as the negative dipole and hydrogen as the positive dipole.

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2 years ago
Will Florine react? will it combine with other elements?
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Explanation:

5 0
3 years ago
Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−
kolbaska11 [484]

Answer:

r = 3.61x10^{-6} M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k.[S2O2^{-8} ]^{x} x [I^{-} ]^{y}

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}

\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y

0.67 = 0.67^x

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}

\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y

0.33 = 0.5x 0.67^y

0.67 = 0.67^y

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

2.6x10^{-6} = kx0.018x0.036kx6.48x10^{-4} = 2.6x10^{-6}

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = 4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}

r = 3.61x10^{-6} M/s

7 0
3 years ago
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