A 0.00143 M concentration of MnO4^- is not a reasonable solution .
<h3>Number of moles of carbonate</h3>
The ions left in solution are Na^+ and NO3^-
Number of moles of calcium nitrate = 100/1000 L × 1 = 0.1 moles
Since;
1 mole of sodium carbonate reacts with 1 mole of calcium nitrate then 0.1 moles of sodium carbonate were used.
<h3>Conductivity of filtrate</h3>
The claim of the student that the concentration of sodium carbonate is too low is wrong because the value was calculated from concentration and volume of calcium nitrate and not using the precipitate. If the filtrate is tested for conductivity, it will be found to conduct electricity because it contains sodium and NO3 ions.
2) In the reaction as shown, the MnO4^- ion was reduced.
The initial volume is 3.4 mL while the final volume is 29.6 mL.
Number of moles of MnO4^- ion = (29.6 mL - 3.4 mL)/1000 × 0.0235 M = 0.0006157 moles
<h3>The calculations are performed as follows</h3>
- If 2 moles of MnO4^- reacted with 5 moles of acid
0.0006157 moles of MnO4^- reacted with 0.0006157 moles × 5 moles/ 2 moles
= 0.0015 moles
- In this case, number of moles of acid = 0.139 g/90 g/mol = 0.0015 moles
Number of moles of MnO4^- = 0.00143 M × (29.6 mL - 3.4 mL)/1000
= 0.000037 moles
- If 2 moles of MnO4^- reacts with 5 moles of acid
0.000037 moles of MnO4^- reacts with 0.000037 moles × 5 moles/ 2 moles
= 0.000093 moles
- Hence, this is not a reasonable amount of solution.
Learn more about MnO4^- : brainly.com/question/10887629
Answer:
Hydroxy(oxo)bismuthine oxide
Answer:
A. 28 years
Explanation:
Applying,
R = R'(2ᵃ/ⁿ).............. Equation 1
Where R = Original sample, R' = Sample left after decay, a = Total time taken to decay, n = half life.
From the question,
Given: R = 12 g, R' = 6 g, a = 28 years.
Substitute into equation 1 and solve for n
12 = 6(2²⁸/ⁿ)
12/6 = 2²⁸/ⁿ
2²⁸/ⁿ = 2
Equation the base,
28/n = 1
n = 28 years.
Hence the half-life is 28 years
Answer: ΔH for the reaction is -277.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28products%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28reactant%29%5D)
![\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCCl_4%7D%5Ctimes%20%5CDelta%20H_%7BCCl_4%7D%29%2B%28n_%7BHCl%7D%5Ctimes%20B.E_%7BHCl%7D%29%20%5D-%5B%28n_%7BCH_4%7D%5Ctimes%20%5CDelta%20H_%7BCH_4%7D%29%2Bn_%7BCl_2%7D%5Ctimes%20%5CDelta%20H_%7BCl_2%7D%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-139%29%2B%281%5Ctimes%20-92.31%29%20%5D-%5B%281%5Ctimes%20-74.87%29%2B%281%5Ctimes%20121.0%5D)
Therefore, the enthalpy change for this reaction is, -277.4 kJ
Emitted/given off. exo is a way to remember that is is going away. think of a flame. the heat is emitted off onto you.
