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3 years ago

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The ion has eight valence electrons. the ion has eight valence electrons. cr6+ ni6+ fe6+ co6+ mn6+

1 answer:

statuscvo [17]3 years ago

7 0

Answer is: Cr⁶⁺.
₂₄Cr 1s²2s²2p⁶3s²3p⁶3d⁵4s¹.
₂₄Cr⁶⁺ 1s²2s²2p⁶3s²3p⁶. 8 valence electrons.
₂₈Ni 1s²2s²2p⁶3s²3p⁶3d⁸4s².
₂₈Ni⁶⁺ 1s²2s²2p⁶3s²3p⁶3d⁴.
₂₆Fe 1s²2s²2p⁶3s²3p⁶3d⁶4s²
₂₆Fe⁶⁺1s²2s²2p⁶3s²3p⁶3d².
₂₇Co 1s²2s²2p⁶3s²3p⁶3d⁷4s²
₂₇Co⁶⁺ 1s²2s²2p⁶3s²3p⁶3d³.
₂₅Mn⁶⁺ 1s²2s²2p⁶3s²3p⁶3d¹.

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RUDIKE [14]

Answer:

The answer is North

Explanation:

The direction of the field is taken to be the direction of the force it would exert on a positive test charge.

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avanturin [10]

B. 11,540

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

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Nt=25 g

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inessss [21]

Answer:

$\Delta S^{0}$ for the given reaction is -99.4 J/K

Explanation:

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$\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]$

where $S^{0}$ represents standard entropy.

Plug in all the standard entropy values from available literature in the above equation:

$\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K$

So, $\Delta S^{0}$ for the given reaction is -99.4 J/K

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What is another name for a beta minus (B-) particle?

pentagon [3]

<span>electron, I believe.
</span>

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