Answer:
MgCO₃
Explanation:
From the question given above, we obtained:
MgF₂ + Li₂CO₃ —> __ + 2LiF
The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:
MgF₂ (aq) —> Mg²⁺ + 2F¯
Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯
MgF₂ + Li₂CO₃ —>
Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:
MgF₂ + Li₂CO₃ —> __ + 2LiF
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Therefore, the missing part of the equation is MgCO₃
The characteristic of the Bohr model that would best support his observation is this assumption: "The energy of the electron in an orbit is proportional to its distance from the nucleus. The further the electron is from the nucleus, the more energy it has." The discrete, bright, colored lines might represent the electrons and its distance from the nucleus. The lights are caused by the energy it has.
Answer: -64.1 kJ.
Explanation:
According to first law of thermodynamics:
=Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done by the system=
{Work is done by the system is negative as the final volume is greater than initial volume}
w = -855 Joules = 0.855 kJ (1kJ=1000J)
q = -65.0 kJ {Heat released by the system is negative}
Thus the change internal energy (ΔE) for a system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings is -64.1 kJ.
