Answer:
Lithium does form a peroxide as well as an oxide on burning in air and I suspect the low temperature reaction with air forms a significant amount of peroxide.
Answer:
The answers to the question are
1. 2nd and above order order
2. 2nd order
3. 1/2 order
4. 1st order
5. 0 order
Explanation:
We have 
1. For nth order reaction half life
∝ ![\frac{1}{[A_{0} ]^{n-1} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%20%5D%5E%7Bn-1%7D%20%7D)
Therefore for a 0 order reaction increasing concentration of the reactant there will increase 
First order reaction is independent [A₀].
Second order reaction [A₀] decrease, increase.
Similarly for a third order reaction
1. 2nd order
2. 2nd order reaction
3. Order of reaction is 1/2.
4. 1st order reaction.
5. Zero order reaction.
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
1. H₂SO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + 2H₂O
2. 2NaOH + H₂CO₃ ⟶ Na₂CO₃ + 2H₂O
3. HNO₃ + KOH ⟶ KNO₃ + H₂O
<em>Explanation</em>:
Acid + base ⟶ salt + water
Take the H from the acid and the OH from the base to get water.
Then, join what’s left to get the salt. Write the symbol for the metal first.
For example, in equation 3, take the H from HNO₃ and the OH from KOH.
Combining the remaining parts (NO₃ and K) to get the salt, KNO₃.
15 is b 16 is b and im working on 14