Respuesta:
14,9 g
Explicación:
Paso 1: Calcular los moles de gas
Una garrafa contiene 9 L de gas a 18°C (291 K) y 780 Torr. Podemos encontrar los moles de gas (n) usando la ecuación del gas ideal.
P × V = n × R × T
n = P × V/R × T
n = 780 torr × 9 L/(62,4 Torr.L/mol.K)/ × 291 K = 0.387 mol
Paso 2: Calcular la masa molar del gas
El gas tiene una densidad (ρ) de 2 g/L a 80 °C (353 K) y 1,5 atm. Podemos calcular la masa molar (M) del gas usando la siguiente fórmula.
ρ = P × M / R × T
M = ρ × R × T / P
M = 2 g/L × (0.082 atm.L/mol.K) × 353 K / 1,5 atm = 38,59 g/mol
Paso 3: Calcular la masa del gas (m)
Usaremos la siguiente formula.
m = n × M = 0.387 mol × 38,59 g/mol = 14,9 g
Answer:please see below for answers in the spaces given.
Explanation:
There are three types of cubic-unit cells of a cubic system which include Simple cubic unit cell, body-centered cubic unit cell and face-centered cubic-unit cell and Thier characteristics are completed below.
1) In the simple cubic unit cell, the centers of _______eight _____ identical particles define the _________corners___ of a cube. The particles do touch along the cube's _______edges_____ but do not touch along the cube's ____diagonal_______ or through the center. There is/are _______one_____ particle per unit cell and the coordination number is
__six______ .
2. In the body-centered cubic unit cell, the centers of _______eight _____ identical particles define the _______corners_____ of the cube plus ______one______ particle at the _______center_____ of ______the cube______ . The particles do not touch along the cube's _______edges_____ or faces but do touch along the cube's ____diagonal________ . There is/are _____two_______ particles per unit cell and the coordination number is _____eight_______ .
3. In the face-centered cubic cell, the centers of ______eight______ identical particles define the _______corner____ of the cube plus ________one____ particle in the _____center_______ of ______each face______ . The particles on the _____corners_______ do not touch each other but do touch those on the ______faces____ . There is/are ________four___ particles per unit cell and the coordination number is _____twelve_______ .
Answer:
1. Electronic configuration of sulphide ion S²¯ (18) => 1s² 2s²2p⁶ 3s²3p⁶
2. Ar
Explanation:
1. S²¯ simply indicates that the sulphur atom (S) has gained two extra electrons.
Thus we shall write electron configuration of sulphur (S) and sulphide ion (S²¯) as follow:
The electronic configuration of S²¯
S (16) => 1s² 2s²2p⁶ 3s²3p⁴
S²¯ (18) => 1s² 2s²2p⁶ 3s²3p⁶
2. Determination of the noble gas element that has the same electronic configuration as S²¯.
This is illustrated below:
Sulphur (S) has 16 electrons while the sulphide ion (S²¯) has extra 2 electrons making it to have a total electron of 18. Observing the elements in the periodic table, only Argon has 18 electrons with electronic configuration as:
Ar (18) => 1s² 2s²2p⁶ 3s²3p⁶
Therefore, sulphide ion (S²¯) has the same electronic configuration as Argon (Ar).
Answer:
32÷5
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