Answer:
a-294
b-3.401×10^-6
c and d - 2.048×10^18
Explanation:
Multiply the RAM for each individual element by its molecular number, e.g., (c-12×14), and then sum to find the molar mass.
b-The molar mass is expressed in grams/mol, hence convert 1 mg to g, which equals 0.001, and divide it by the molar mass.
c/d-1 mole of any substance consists of 6.023×10^23 (ions, molecules, etc.), therefore we need to find the moles here as
(3.401×10^-6) × (6.023×10^23).
Answer:
2.5 g of platinum
Explanation:
A catalyst is a substance added to a reaction to enhance the reaction speed. It does not undergo any change during the reaction, meaning it remains unchanged after the reaction concludes. The role of a catalyst is to provide an alternative pathway for the reaction by reducing the activation energy required. Therefore, a catalyzed reaction occurs more rapidly and requires less energy compared to an uncatalyzed one.
Since catalysts do not get involved in reactions and retain their mass post-reaction, the amount of platinum will stay the same (2.5g). The mass can only alter if a substance participates in the chemical process. Thus, this is the response.
Cu(NO3)2 --> MM187.5558
NiNO3 *COEF2* --> 120.6983
The appropriate answer is option E. Gibbs free energy can be expressed using the equation: ΔG = ΔH - TΔS, where ΔH denotes the change in enthalpy of the reaction, T is the reaction temperature, and ΔS signifies entropy change. For our calculations, we have ΔH = -720.5 kJ/mol which converts to -720500 J/mol (given that 1 kJ = 1000 J), ΔS = -263.7 J/K, and T = 141.0°C, which equals 414.15 K. Consequently, the Gibbs free energy for the specified reaction at 141.0°C is calculated as -611.3 kJ/mol.
The enthalpy change in this scenario totals 7.205 KJ. The task is to compute the enthalpy variation during the conversion of 10.0 g of ice at -25.0°C into water at 80.0°C, factoring in specific heats and enthalpy for phase transitions.
