The pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.
Explanation:
In the present problem, the temperature is said to remain at constant and there is change in the pressure. So according to Boyle's law, the relationship between pressure and volume of any gaseous objects are inversely related to each other. In other words, the pressure attained by gas molecules in a container will be inversely proportional to the volume of the gas molecules occupied in the container, at constant temperature.
So, if two volumes V₁ and V₂ are considered, then their respective pressure will be represented as P₁ and P₂. Then, as per Boyle's law,
So let us consider, V₁ = 6 cm³ and V₂ = 4 cm³ and pressure P₁ = 405 kPa and we have to determine P₂.
Then, 
So, the pressure at new volume of 4 cm³ is 486 kPa. It can be seen that as there is decrease in the volume, there is an increase in the pressure. So it satisfied the Boyle's law.
Thus, the pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.
The correct answer would be equation 4, because there are 4 Hydrogens and 2 Oxygens on each side of the equation making it balanced.
Answer:
ZnS(s) ⇄ S²⁻(aq) + Zn²⁺(aq)
Explanation:
First, we will write the molecular equation, since it is easier to balance.
2 HBr(aq) + ZnS(s) ⇄ H₂S(aq) + ZnBr₂(aq)
In the full ionic equation we include all ions and molecular species.
2 H⁺(aq) + 2 Br⁻(aq) + ZnS(s) ⇄ 2 H⁺(aq) + S²⁻(aq) + Zn²⁺(aq) + 2 Br⁻(aq)
In the net ionic equation we include only the ions that participate in the reaction and the molecular species.
ZnS(s) ⇄ S²⁻(aq) + Zn²⁺(aq)
Answer:
2 C Atoms
Explanation:
When you have coefficient of 2 next to a compound element, it indicates there are 2 of each compound element. In the compound element, there is one C Atom, and 2×1 is 2.
Answer:
Explanation:
Dado que:
masa de oxígeno gaseoso = 100 g
presión = 1 atm
temperatura = 273 K
(a)
número de moles de oxígeno contenidos en el matraz = masa de oxígeno / masa molar de oxígeno
= 100 g / 16 gmol⁻¹
= 6.25 moles
(b) El número de moléculas de oxígeno es el siguiente:
Dado que 1 mol de oxígeno gaseoso contiene 6.023 * 10²³ moléculas de oxígeno.
Entonces, 6.25 moles contendrán:
= (6.25 × 6.023 * 10²³) moléculas de oxígeno.
≅ 3.764 × 10²³ moléculas de oxígeno.
(c) El número de átomos de oxígeno es:
= 2 × 3.764 × 10²³
= 7.528 × 10²³ átomos de oxígeno
(d) Usando la ecuación de gas ideal
PV = nRT
El volumen ocupado por el oxígeno = 
Volumen ocupado por oxígeno = 
Volumen ocupado por oxígeno= 14185.76 m³
