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2 years ago

Write an equation that shows the formation of the phosphide ion from a neutral phosphorus atom.

Chemistry

2 answers:

atroni [7]2 years ago

8 0

Answer: $P+3e^-\rightarrow P^{3-}$

Explanation: Phosphorous atom is the element with atomic number 15 and thus contains 15 electrons.

$P:15:1s^22s^22p^63s^23p^3$

Thus it is short of 3 electrons to attain stable noble gas configuration. Thus it gains three electrons and converts to phosphide ion $(P^{3-})$ with 18 electrons.

$P^{3-}:18:1s^22s^22p^63s^23p^6$

Vlad1618 [11]2 years ago

5 0

From reference or science books, we can actually know that the elementary charge of phosphorus is -3. Since it is negative, this means that it has an excess of electrons, which further means that it takes up 3 electrons to become an ion. Therefore the equation can be written as:

P + 3e ---> P3-

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You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number

irina1246 [14]

Answer: $9.11\times 10^{-31}kg$

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus $9.11\times 10^{-31}kg$ has three significant figures

7 0

2 years ago

What type of bond joins the carbon atom to each of the hydrogen atoms? A molecule that consists of four hydrogen atoms and one c

Len [333]

Answer:

Explanation:

what is the questions??

4 0

2 years ago

Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4

Ghella [55]

Answer:

The K sp Value is $K_{sp}=7.40$

Explanation:

From the question we are told that

The of $KClO_3$ is = 122.5 g/ mol

The mass of $KClO_3$ dissolved is $m_s = 4.0g$

The volume of solution is $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

$No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

$= \frac{4}{122.5}$

$=0.0327\ moles$

Generally concentration is mathematically represented as

$concentration = \frac{No \ of \ moles}{volume }$

For $KClO_3$

$Z= \frac{0.0327}{12*10^{-3}}$

$=2.72 \ mol/L$

The dissociation reaction of $KClO_3$ is

$KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}$

The solubility product constant is mathematically represented as

$K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }$

Since there is no ionic reactant we have

$K_{sp} = [k^+] [ClO_3^-]$

$= Z^2$

$= 2.72^2$

$K_{sp}=7.40$

5 0

2 years ago

Read 2 more answers

calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c

Pepsi [2]

Answer:

The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)

Explanation:

We make use of the formula

$Q=m \times c \times \Delta T$

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

$Q=m \times c \times \Delta T$

Rearranging the formula

$c= \frac {Q}{(m\times \Delta T)}$

$= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}$

So,

c = 0.129 J/(g℃)

(Answer)

7 0

2 years ago

A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is added to the mi

vekshin1

Answer:

Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

Explanation:

The equilibriums that take place are:

Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

Co⁺² + 2OH⁻ ↔ Co(OH)₂(s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

Keep in mind that the concentration of each ion is halved because of the dilution when mixing the solutions.

For Cu⁺²:

2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

2.2x10⁻²⁰ = 0.25 M*[OH⁻]²

[OH⁻] = 2.97x10⁻¹⁰ M

For Co⁺²:

1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²

[OH⁻] = 7.21x10⁻⁸ M

Because Copper requires less concentration of OH⁻ than Cobalt, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

3 0

2 years ago