Answer:
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Explanation:
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In this case, since the integrated rate law for a second-order reaction is:
![[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B%5BSO_3%5D_0%7D%7B1%2Bkt%5BSO_3%5D_0%7D)
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
![[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B1.44M%7D%7B1%2B14.1M%5E%7B-1%7Ds%5E%7B-1%7D%2A0.240s%2A1.44M%7D%5C%5C%5C%5C)
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<u>Answer:</u> The red litmus paper turns blue on dipping in NaOH solution.
<u>Explanation:</u>
Litmus paper is the indicator that detects the nature of the solution, whether it is acidic or basic.
There are 2 types of litmus paper:
- <u>Red litmus paper:</u> This paper will turn blue if it is dipped in basic solution and will remain as such if it is dipped in acidic solution.
- <u>Blue litmus paper:</u> This paper will turn red if it is dipped in acidic solution and will remain as such if it is dipped in basic solution.
NaOH is a strong base, so when a red litmus paper is dipped in the beaker having necessary amount of NaOH, the red litmus paper turns into blue.
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:
Noble Gases
Explanation:It's not on there but Noble gases is the least reactive element because it has 8 electrons and their outer energy is full.
