Which diagram shows the correct way to represent an ionic compound of magnesium oxide?

Janine is trying to separate some ethanol from water. Which method should she use does anyone know this

GuDViN [60]

To separate some ethanol from water, Jane needs to use the method of simple distillation.

Simple distillation is a separation method in which two liquids can be separated from each other based on differences in boiling point.

Since water has a higher boiling point than ethanol, ethanol is collected first as the distillation proceeds.

Learn more: brainly.com/question/2193327

5 0

2 years ago

Consider these generic half-reactions. Half-reaction E° (V) X+(aq)+e−⟶X(s) 1.52 Y2+(aq)+2e−⟶Y(s) −1.17 Z3+(aq)+3e−⟶Z(s) 0.84 Ide

olya-2409 [2.1K]

Answer:

strongest oxidizing agent: $X^{+}$

weakest oxidizing agent: $Y^{2+}$

strongest reducing agent: Y

weakest reducing agent: X

$X^{+}$ will oxidize Z

Explanation:

The higher the reduction potential of a species, higher will be the tendency to consume electrons from another species. Hence higher will be the oxidizing power of it's oxidized form and lower will be the reducing power of it's reduced form.

Alternatively, higher reduction potential value suggests that the oxidized form of the species acts as a stronger oxidizing agent and the reduced form of the species acts as a weaker reducing agent.

Order of reduction potential:

$E_{X^{+}\mid X}^{0}(1.52V)> E_{Z^{3+}\mid Z}^{0}(0.84V)> E_{Y^{2+}\mid Y}^{0}(-1.17V)$

So, strongest oxidizing agent: $X^{+}$

weakest oxidizing agent: $Y^{2+}$

strongest reducing agent: Y

weakest reducing agent: X

As reduction potential of the half cell $X^{+}\mid X$ is higher than the reduction potential of the half cell $Z^{3+}\mid Z$ therefore $X^{+}$ will oxidize Z into $Z^{3+}$ and itself gets converted into X.

5 0

2 years ago

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −

Grace [21]

Answer : The value of $K_{eq}$ is, 11.2

The value of $\Delta G_{rxn}$ is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$\Delta G^o=-RT\times \ln K_{eq}$

$-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=11.2$

Thus, the value of $K_{eq}$ is, 11.2

Now we have to calculate the $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

The given reaction is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

$\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})$

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -9.04 kJ/mol

3 0

2 years ago

Find the ions in the periodic table that have an electron configuration of nd8 (n = 3, 4, 5...).

Alla [95]

<span>Cu⁺ is the only one of the ions in the list that will show 8 electrons in a d sublevel....its configuration will be Ar| 4s² 3d⁸
hope this helps</span>

4 0

2 years ago

Read 2 more answers

Consider a saturated solution formed when 17.5 g of a solute dissolve in 28.3 g of a solvent, giving a total solution volume of

STALIN [3.7K]

Answer:

a) 38.2 % mass

b) 61.8 g solute/100 g solvent

c) 1.65 g/mL

Explanation:

Given the data:

mass of solute = 17.5 g

mass of solvent= 28.3 g

total solution volume= 27.8 mL

a)- mass percent= mass of solute/mass of solution x 100

mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g

mass % = 17.5 g/45.8 g x 100 = 38.2 % mass

b)- solubility = grams of solute/ 100 g solvent

= 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent

c)- density = massof solution/total volumesolution = 45.8 g/27.8 mL = 1.65 g/mL

7 0

2 years ago