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
Answer:
Oxidation of potassium amalgam with carbon dioxide results in the formation of potassium oxalate. Potassium is not reactive with benzene, although heavier alkali metals such as cesium react to give organometallic products.
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Answer: Ti is the reducing agent because it changes from 0 to +4 oxidation state.
Explanation:
- Firstly, we need to identify the reducing agent and the oxidizing agent.
- The reducing agent: is the agent that has been oxidized via losing electrons.
- The oxidizing agent: is the agent that has been reduced via gaining electrons.
- Here, Ti losses 4 electrons and its oxidation state is changed from 0 to +4 and Cl₂ gains one electron and its oxidation state is changed from 0 to -1.
- So, Ti is the reducing agent because its oxidation state changes from 0 to +4.
- Cl₂ is the oxidizing agent because its oxidation state changes from 0 to -1.
- Thus, The right answer is Ti is the reducing agent because it changes from 0 to +4 oxidation state.
1,363 is the right answer
It would be C because you would do m divided by s squared and you would get 0.11 but you would round it to 0.12
