<span>The partial pressure of oxygen is 438.0 mmHg. The ideal gas equation is expressed as PV = nRT where P represents pressure, V denotes volume, n is the number of moles, R is the ideal gas constant (8.3144598 (L*kPa)/(K*mol)), and T signifies absolute temperature. To convert from Celsius to Kelvin, we have 43.4 + 273.15 = 316.55 K. For the pressure conversion from mmHg to kPa: 675.9 mmHg * 0.133322387415 = 90.11260165 kPa. When solving for n using the ideal gas equation, we derive n = PV / (RT) which provides n = 90.11260165 kPa * 16.2 L / (8.3144598 (L*kPa)/(K*mol) * 316.55 K)= 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol), resulting in n = 0.554656603 mol. Thus, we have 0.554656603 moles of gas particles. Next, we determine the contribution from oxygen. The atomic weight of oxygen is 15.999 g/mol, while argon is 39.948 g/mol, and the molar mass of O2 is 31.998 g/mol. We establish the relationships where M is the number of moles of O2, and 0.554656603 - M gives the number of moles of Ar. Setting up the equation: M * 31.998 + (0.554656603 - M) * 39.948 = 19.3, we solve for M resulting in 0.359424148 moles of oxygen out of 0.554656603 total moles. This leads to oxygen providing 0.359424148 / 0.554656603 = 0.648012024 or 64.8012024% of the total pressure of 675.9 mmHg. The partial pressure therefore calculates to 675.9 * 0.648012024 = 437.9913271 mmHg, rounded to 438.0 mmHg</span>
<span>Salts result from the reaction between bases and water. - FALSE
</span><span>Most salts are ionic and dissolve in water. - TRUE
</span><span>Most salts are not dissolved in water and do not have electrical charges. - FALSE
</span><span>Solutions containing salt and water are unable to conduct electricity. - FALSE
:)</span>
Respuesta:
0.16 M
Explicación:
Teniendo en cuenta:
O sea,
Dado que:
Para 
:
Molaridad = 0.200 M
Volumen = 20.0 mL
Convierte mL a L:
1 mL = 10⁻³ L
Entonces, volumen = 20.0×10⁻³ L
Los moles de son:
Moles de = 0.004 moles
Para 
:
Molaridad = 0.400 M
Volumen = 30.0 mL
Convertimos mL a L:
1 mL = 10⁻³ L
Volumen = 30.0×10⁻³ L
Entonces, los moles de son:
Moles de = 0.012 moles
Según la reacción:
1 mol de reacciona con 1 mol de
Por lo tanto,
0.012 mol de reacciona con 0.012 mol de
Moles disponibles de = 0.004 mol
El reactivo limitante es el que está en menor cantidad, entonces es el limitante (0.004 < 0.012).
La formación del producto depende del reactivo limitante, así que,
1 mol de reacciona con 1 mol de y produce 1 mol de 
0.004 mol de reacciona con 0.004 mol de y genera 0.004 mol de
Los moles restantes de son: 0.012 - 0.004 = 0.008 mol
El volumen total es 20 + 30 mL = 50 mL = 0.050 L
Por lo que la concentración del ion bario, 
, después de la reacción es:
To achieve the cancellation of electrons, the oxidation half-reaction needs to be multiplied by 4 while the reduction half-reaction must be multiplied by 3. Explanation: The oxidation reaction accounts for the loss of electrons, increasing the oxidation state, while the reduction implies gaining electrons, leading to a decrease in oxidation state. The respective half-reactions illustrate this, confirming that multiplying the oxidation by 4 and the reduction by 3 achieves the desired effect.
