Difference between Gas and Vapour:
Gas:
A thermodynamic state in which a substance exists only in one phase i.e. Gas phase. In above given examples N₂, He and CO₂ exists as gases at room temperature. These gases cannot form a solid or liquid phase along with gas phase as these states requires very low temperatures.
Vapours:
It is a thermodyanamic state in which a substance exists in more than one phase. In given options Sulfur can exist in vapor state. This is because sulfur exists in a cyclic or chain form due to catenation (self linkage property). Therefore, a lower members of S allotrops can form a vapours.
Given what we know, we can confirm that in a voltaic cell, the anode loses electrons and is oxidized, meanwhile, the cathode is reduced by gaining electrons.
<h3 /><h3>What is a voltaic cell?</h3>
- It is described as an electrochemical cell.
- These cells use chemical reactions to produce electrical energy.
- During this reaction, an anode loses electrons, thus oxidizing.
- Meanwhile, the cathode gains electrons and is reduced.
Therefore, given the nature of the voltaic cell, we can confirm that during its reaction, the anode is oxidized by losing electrons while the cathode becomes reduced by gaining them.
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Answer:
Substance is Aluminium
Explanation:
We are given;
Mass; m = 13.5 grams
Volume; V = 5 cm³
Formula for density is;
density = m/V
Density = 13.5/5
Density = 2.7 g/cm³
From the table attached, we can see that the element with a corresponding density of 2.7 is Aluminium
Answer:
s = 4.41 g/L.
Explanation:
¡Hola!
En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:
Lo cual hace que la expresión de equilibrio se calcule como:
![Ksp=[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)
Y que en términos de la solubilidad molar, s, se resuelve como:
![1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L](https://tex.z-dn.net/?f=1.6x10%5E%7B-5%7D%3Ds%282s%29%5E2%5C%5C%5C%5C1.6x10%5E%7B-5%7D%3D4s%5E3%5C%5C%5C%5Cs%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1.6x10%5E%7B-5%7D%7D%7B4%7D%20%7D%20%5C%5C%5C%5Cs%3D0.0159molPbCl_2%2FL)
Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):
¡Saludos!
Given:
P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?
Assume that the gas ideally, thus, we can use the ideal gas equation:
PV = nRT
where R = 0.0821 L atm/mol K
123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T
solve for T
T = 295.72 K<span />
