Balance the following equation. __AI+ __CuO __AI203+

When the oxide of generic metal m is heated at 25.0 °c, only a negligible amount of m is produced. o2m(s ----> m(s o2(g delta

Len [333]

(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K = $e^{ \frac{-dG0}{RT} }$
= $e^{ \frac{105500}{8.314 x 298.15} }$
= 3.05 x 10¹⁸

6 0

3 years ago

What mass of magnesium chloride is needed to make 100.0 mL of a solution that is 0.500 M in chloride ion?

miss Akunina [59]

M = n/V

.5M = n/.100 L

n = .1 L * .5M

n= .05 mols of MgCl2

mass of MgCl2 = .05 mols of MgCl2 * 95.211 grams/ 1 mol of MgCl2

mass of MgCl2 = 4.76 grams

4.76 grams of MgCl2 is needed to make 100 ml of a solution that is .500M, in chloride ion. Bolded = confused

3 0

3 years ago

What are typical characteristics of metals

photoshop1234 [79]

Answer:

Metals are lustrous, malleable, ductile, good conductors of heat and electricity. Other properties include: State: Metals are solids at room temperature with the exception of mercury, which is liquid at room temperature (Gallium is liquid on hot days).

4 0

3 years ago

If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?

olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

HA H₃O⁺ A⁻

Initial, M 0.40 0 0

Change , M -x +x +x

Equilibrium, M 0.40 - x x x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0

3 years ago

HELP PLEASE THE OTHER 'ANSWER' ISNT EVEN AN ANSWER!

hodyreva [135]

Answer:

most likely that (2) the replicated experiment was performed incorrectly.

Why, u ask? u dare question me:

1- The initial experiment invalidness cannot be proven.

2- the second answer is correct

3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.

4- Already knowing the data and errors would increase the precision of the replicated experiment.

5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.

happy learning!

4 0

3 years ago

Balance the following equation. __AI+ __CuO __AI203+ _Cu

Balance the following equation. AI+ CuO __AI203+ _Cu