Which of the following is a small-scale, sudden natural change?

What trend in ionization do you see going up and down?

xxTIMURxx [149]

Depending if you were going up, you would be seeing an increase in relative ionization energies in a particular group on the periodic table, and vice versa, if you go down the group, an decrease in the relative ionization energies.

5 0

3 years ago

¿Qué son los lantánidos y actínidos?

harkovskaia [24]

Answer: En estos grupos de los elementos de transición se encuentran las llamadas tierras raras, separadas del resto de elementos de la tabla, que pertenecen al grupo IIIB y se les conoce como lantánidos y actínidos.

6 0

2 years ago

A fuel was burned for 5 min, increasing the temperature of 10.0 g of water with a density of 1.00 g/ml by 9.0 oC. The fuel relea

jekas [21]

The fuel released 90 calories of heat.

Let suppose that water experiments an entirely sensible heating. Hence, the heat released by the fuel is equal to the heat absorbed by the water because of principle of energy conservation. The heat released by the fuel is expressed by the following formula:

$Q = m\cdot c \cdot \Delta T$ (1)

Where:

$m$ - Mass of the sample, in grams.
$c$ - Specific heat of water, in calories per gram-degree Celsius.
$\Delta T$ - Temperature change, in degrees Celsius.

If we know that $m = 10\,g$ , $c = 1\,\frac{cal}{g\cdot ^{\circ}C}$ and $\Delta T = 9\,^{\circ}C$ , then the heat released by the fuel is:

$Q = (10\,g)\cdot \left(1\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (9\,^{\circ}C)$

The fuel released 90 calories of heat.

We kindly invite to check this question on sensible heat: brainly.com/question/11325154

7 0

2 years ago

What is the wavenumber of the radiation emitted when a hydrogen

LUCKY_DIMON [66]

Answer: Wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

$E=\frac{hc}{\lambda}$

where,

E = energy of the radiation = $1.634\times 10^{-18}J$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of radiation = ?

Putting values in above equation, we get:

$1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m$

$\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}$

Thus wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

8 0

4 years ago

A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a

Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

BaC₂O₄ ⇌ Ba²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 5.0 × 10⁻⁵ c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

ZnC₂O₄ ⇌ Zn²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 2.0 × 10⁻⁷ c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻

E/mol·L⁻¹: 3.0 × 10⁻⁵ c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028 mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0

3 years ago