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Valentin [98]
2 years ago
13

In the writing of ionic chemical formulas, what factor is "crossed over" in the crossover rule?

Physics
1 answer:
Aleks04 [339]2 years ago
3 0

In the writing of ionic chemical formulas the value of each ion's charge is crossed over in the crossover rule.

Rules for naming Ionic compounds

  • Frist Rule
    The cation (element with a negative charge) is written first in the name then the anion(element with a positive charge) is written second in the name.
  • Second rule
    When the formula unit contains two or more of the same polyatomic ion, that ion is written in parentheses with the subscript written outside the parentheses.
    Example: Sodium carbonate is written as Na₂CO₃ not Na₂(CO)₃
  • Third rule
    If the cation is a metal ion with a fixed charge then the name of the cation will remain the same as the (neutral) element from which it is derived (Example: Na+ will be sodium).
    If the cation is a metal ion with a variable charge, the charge on the cation is indicated using a Roman numeral, in parentheses, immediately following the name of the cation (example: Fe³⁺ = iron(III)).
  • Fourth rule
    If the anion is a monatomic ion, the anion is named by adding the suffix <em>-ide</em> to the root of the element name (example: F = Fluoride).

The oxidation state of each ion is also important, thus in the crossover rule, the value of each ion's charge is crossed over.

Learn more about chemical formulas here:

<u>brainly.com/question/11995171</u>

#SPJ4

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All stars in a stellar cluster have roughly the same:___.
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All stars in a stellar cluster have roughly the same distance.

<h3>What coloration are celebrity clusters?</h3>

Open clusters have a tendency to be blue in color. They frequently include glowing gas and dust. The stars in an open cluster are young stars that all formed from the equal nebula. These warm blue stars are in an open cluster known as the Jewel Bo

<h3>Are stars in the identical cluster?</h3>

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4 0
1 year ago
The three factors that determine the amount of potential energy in an object are?
ozzi

Answer:

ggy h Jr scythe fund the CT h hytgy6fhhj

4 0
2 years ago
Two cars go through 2 different crashes. Car 1 experiences a 500 Ns impulse for a duration of 15s, while car 2 experiences that
tatuchka [14]
B is your answer hope this helps;)
5 0
3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
grin007 [14]

Answer:

f_e = 1.51 cm

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}

400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}

9.6 = \dfrac{16-f_e}{f_e}

9.6f_e = 16-f_e

10.6 f_e = 16

f_e = 1.51 cm

3 0
3 years ago
A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if
Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

F_p = 6*10^{22}N

F_p = \frac{GMm}{R^2}

F_p = 9*10^{22}N

Distance between planet and star

r = \frac{R}{2}

Gravitational force is

F = \frac{GMm}{r^2}

Applying the new distance,

F = \frac{GMm}{(\frac{R}{2})^2}

F =  4\frac{GMm}{R^2}

Replacing with the previous force,

F = 4F_p

Replacing our values

F= 4(9*10^{22}N)

F = 36*10^{22}N

Therefore the magnitude of the force on the star due to the planet is  36*10^{22}N

5 0
3 years ago
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