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Softa [21]
3 years ago
7

Can you write the formula for and indicate the charge on each of the following ions: sodium ion, aluminum ion, chloride ion, nit

ride ion, iron(11) ion, iron(111) ion
Chemistry
1 answer:
Anettt [7]3 years ago
8 0
If the ion is named by (element-name) + ion, it is the element with a positive charge.
This means:
Sodium ion: Na(1+)
Aluminum ion: Al(3+)
If the ion is named by (element-prefix)+ide, then it has a negative charge.
This means:
Chloride: Cl(1-)
Nitride: N(3-)
Finally, if it has roman numerals after the name, that is the amount of positive charge the ion has.
Therefore:
Iron (II): Fe(2+)
Iron(III): Fe(3+)
You might be interested in
Eugen Goldstein discovered in 1886 that atoms also have _________ charges.
Dvinal [7]
The correct answer for the question that is being presented above is this one: "b. positive" Eugen Goldstein discovered in 1886 that atoms also have positive charges. He was a German physicist and an early investigator of discharge tubes. He also discovered the anode rays.

7 0
3 years ago
A 1.0 liter flask contains 90.1 moles of water vapor at 27.0 oC. What is the pressure of the gas?
zmey [24]

Answer:

The pressure of the gas is 224839.8 atm

Explanation:

As we know

PV = nRT

Substituting the given values, we get -

P * 1 L = 90.1 moles * 8.314 4621(75). J K−1 mol−1 * 300

P = 224839.8 atm

The pressure of the gas is 224839.8 atm

5 0
3 years ago
Imagine a solution containing only water, sodium, and glucose. if a voltage is applied to the solution, glucose would move
andrew-mc [135]

There would be no net movement of glucose

The given statement is wrong. Glucose solutions are not electrolytes and therefore do not conduct electricity.

Is sodium chloride a good conductor of electricity? Simply by dissolving or dissolving it in water, it cannot conduct electricity as a solid.

Saline contains Na+ and Cl- ions. If ions are present in the solution, these ions can move through the solution and carry an electric current. Pure water and sugar solutions do not contain significant ions. Sugar molecules do not break down into ions when dissolved.

When glucose is added to water, it dissolves readily but does not dissociate into ions. Therefore, a glucose solution is not an electrolyte and is not a conductor of electricity.

Substances that dissolve in water to form a conductive solution are called electrolytes. Substances that dissolve in non-conductive solutions are called non-electrolytes. All soluble ionic compounds are electrolytes.

Learn more about glucose here;

brainly.com/question/461228

#SPJ4

3 0
2 years ago
Hi! I wanted to know what my grade would be for my test. There are 23 questions in total. 2 of the 23 questions are essay questi
kolbaska11 [484]

Answer:

high B or a low A:) Good Luck

Explanation:

6 0
3 years ago
H2 (g) + I2 (g) ⇌ 2 HI(g) Kc = 54.3 at 430 °C . What will be the concentrations of all species at equilibrium at this temperatur
weqwewe [10]

Answer:

[H₂]  = 6.74×10⁻³ M

[I₂]  = 4.65×10⁻³ M

[HI] = 0.0413 M

Explanation:

The equilibrium is:

H₂(g) + I₂(g) ⇌ 2 HI(g)

Expression for Kc = [HI]² / [H₂] .  [I₂]            54.3

We analyse the equation:

                 H₂(g)      +     I₂(g)      ⇌      2 HI(g)

Initially   0.00623      0.00414              0.0424

React            x                  x                        2x

x amount has reacted, therefore by stoichiometry 2x has been added to the HI, that we have in the beginning.

Eq         0.00623-x    0.00414-x             0.0424+2x

We make the expression for Kc:

Kc = (0.0424+2x)² / (0.00623-x) . (0.00414-x)  = 54.3

This is quadractic funcion:

54.3 = (1.79×10⁻³ + 0.1696x + 4x²) / (2.58×10⁻⁵- 0.01037x + x²)          

54.3 (2.58×10⁻⁵- 0.01037x + x²) = 1.79×10⁻³ + 0.1696x + 4x²

1.40×10⁻³- 0.563x + 54.3x² = 1.79×10⁻³ + 0.1696x + 4x²

-3.89×10⁻⁴ - 0.7326x +50.3x² = 0 → a = 50.3 ; b= - 0.7326 ; c = -3.89×10⁻⁴

Quadratic formula = (-b +- √(b² + 4ac))/ (2a)

x₁ = 0.015

x₂ = -5.13×10⁻⁴  . We choose x₂ as x₁ give us negative concentrations

[H₂] = 0.00623 - (-5.13×10⁻⁴) = 6.74×10⁻³ M

[I₂] = 0.00414 - (-5.13×10⁻⁴) = 4.65×10⁻ M

[HI] = 0.0424+2(-5.13×10⁻⁴) = 0.0413 M

8 0
3 years ago
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