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muminat
3 years ago
8

Describe how to write an ion

Chemistry
2 answers:
Papessa [141]3 years ago
6 0

Answer:

When writing an ion, state symbols of the substances must be clearly indicated.I HOPE IT HELPS I MADE AN EXAMPLE TOO:)

Explanation:

<u>Example: </u>

<u> Write the ionic equation for the word equation </u>

Sodium chloride(aq) + silver nitrate(aq) → silver chloride(s) + sodium nitrate(aq)

<u> Solution: </u>

<u> Step 1: </u>Write the equation and balance it if necessary

NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

<u>Step 2: </u>Split the ions. (Only compounds that are aqueous are split into ions.)

Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) → AgCl(s) + Na+(aq) + NO3-

<u>Step 3: </u>Cancel out spectator ions. (Spectator ions are ions that remain the same in their original states before and after a chemical reaction.)

<u> Ionic Equation </u>

<u>Step 4: </u>Write a balanced ionic equation

Ag+(aq) + Cl-(aq) → AgCl(s)

IrinaVladis [17]3 years ago
5 0

Answer:

Cations (positively-charged): With a superscript + (e.g. \mbox{Na^{+}})

Anions (negative-charged): With a superscript - (e.g. \mbox{Cl^{-}})

Explanation:

An <em>ion</em> is an atom that carries some sort of negative or positive charge due to losing or gaining electrons in bonding with another atom. A <em>cation</em>, a positively-charged ion, loses an electron in the bonding process, and is written with a small plus sign superscript, while an <em>anion</em>, a negatively-charged ion, is written with a small minus sign. For example, a sodium cation in a sodium chloride molecule would be written like \mbox{Na^{+}}, and the corresponding chloride anion would be written like \mbox{Cl^{-}})

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Please help please with the answers
eimsori [14]

1) We know that rtp = number of moles × 24

                                 = 0.25 × 24

                                 = 6 dm³

Therefore the volume of 0.25 moles of gas at rtp is A) 6 dm³

3.1) Amount of Copper = 20 tonnes

Amount of pure copper from impure copper = 18 tonnes

Purity of copper = (Pure copper/ impure copper) × 100

                           = (18 / 20) × 100

                           = 18 × 5

                           = 90 %

Therefore the purity of copper is 90%

3.2) We know that oxygen has 8 protons and 8 neutrons, so the weight of oxygen molecule is 8 + 8 = 16 u

So, one mole of oxygen weighs 16 g, so 2 moles weigh 2 × 16 = 32 grams

But oxygen is diatomic so the weight is 32 × 2 = 64 grams

Therefore the weight of 2 moles of oxygen is 64 grams

3.3) Concentration of solution = Amount of solute/volume

                                                   = 4 moles/ 2 dm³

                                                   = 4 / 2

                                                   = 2 mol/dm³

Therefore the concentration of a solution containing 4 moles in 2 dm³ of solution is 2 mol/dm³

Happy to help :)

If you need any more help, fell free to ask

Extremely sorry for the late answer

7 0
3 years ago
The equilibrium constant, K, for the following reaction is 2.44×10-2 at 518 K: PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture o
Volgvan

Answer:

[PCl₅] = 0.5646M

[PCl₃] = 0.1174M

[Cl₂] = 0.1174M

Explanation:

In the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

K equilibrium is defined as:

<em>K = 2.44x10⁻² = [PCl₃] [Cl₂] / [PCl₅]</em>

The initial moles of each compound when volume is 15.3L are:

PCl₅ = 0.300mol/L×15.3L = 4.59mol

Cl₂ = 8.55x10⁻²mol/L×15.3L = 1.308mol

PCl₃ = 8.55x10⁻²mol/L×15.3L = 1.308mol

At 8.64L, the new concentrations are:

[PCl₅] = 4.59mol / 8.64L = 0.531M

[PCl₃] = 1.308mol / 8.64L = 0.151M

[Cl₂] = 1.308mol / 8.64L = 0.151M

At these conditions, reaction quotient, Q, is:

Q = [0.151M] [0.151M] / [0.531M]

Q = 4.29x10⁻²

As Q > K, <em>the reaction will shift to the left producing more reactant, </em>that means equilibrium concentrations are:

[PCl₅] = 0.531M + X

[PCl₃] = 0.151M - X

[Cl₂] = 0.151M - X

<em>Where X is reaction coordinate.</em>

Replacing in K expression:

2.44x10⁻² = [0.151M - X] [0.151M - X] / [0.531M + X]

1.296x10⁻² + 2.44x10⁻²X = 0.0228 - 0.302X + X²

<em>0 = 9.84x10⁻³ - 0.3264X + X²</em>

Solving for X:

X = 0.293 → False solution. Produce negative concentrations

<em>X = 0.0336M → Right solution.</em>

Replacing:

[PCl₅] = 0.531M + 0.0336

[PCl₃] = 0.151M - 0.0336

[Cl₂] = 0.151M - 0.0336

<h3>[PCl₅] = 0.5646M</h3><h3>[PCl₃] = 0.1174M</h3><h3>[Cl₂] = 0.1174M</h3>
4 0
4 years ago
2<br> 1. Biology is the study of
Inessa05 [86]
Biology is the study of life
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4 years ago
g When copper(II) chloride and sodium carbonate solutions are combined, solid copper(II) carbonate precipitates, leaving a solut
Elden [556K]

Answer:

When copper(II) chloride and sodium carbonate solutions are combined, solid copper(II) carbonate precipitates, leaving a solution of sodium chloride. Write the conventional equation, total ionic equation, and net ionic equation for this reaction.

Explanation:

The word equation for the reaction is:

Copper (II) chloride(aq) + sodium carbonate (aq) ->sodium chloride (aq) +           copper carbonate(s)

The balanced chemical equation of the reaction is:

CuCl_2(aq)+Na_2CO_3(aq)->2NaCl(aq)+CuCO_3(s)

The complete ionic equation is:

Cu^2+(aq)+2Cl^-(aq)+2Na^+(aq)+CO_3^2^-(aq)->2Cl^-(aq)+2Na^+(aq)+CuCO_3(s)\\

The net ionic equation is obtained from the complete ionic equation after removing the spectator ions:

Cu^2^+(aq)+CO_3^2^-(aq)->CuCO_3(s)

5 0
3 years ago
It is true/falseit that when hydrogen cyanide, h c n , dissolves in water, the solution is weakly conducting and acidic in natur
cluponka [151]
False is the correct answer.
8 0
3 years ago
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