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
The percentage by mass of Na2CO3 in the sample is 48%.
The equation of the reaction of Na2CO3 with HCl;
Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)
Since the HCl is in excess, the excess is back titrated with NaOH as follows;
NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)
Number of moles of HCl added = 0.100 M × 50/1000 L = 0.005 moles
Number of moles of NaOH added = 5.6/1000 × 0.100 M = 0.00056 moles
Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.
Amount of HCl that reacted with Na2CO3 = 0.005 moles - 0.00056 moles = 0.0044 moles
Now;
1 mole of Na2CO3 reacts with 2 moles of HCl
x moles of Na2CO3 reacts with 0.0044 moles of HCl
x = 1 mole × 0.0044 moles / 2 moles
x = 0.0022 moles
Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g
Percentage of Na2CO3 in the sample = 0.24 g/ 0.500-g × 100/1 = 48%
Learn more about back titration: brainly.com/question/25485091
Answer:
3m/s²
Explanation:
Given parameters:
Mass of object = 3.2kg
Force to the right = 16.3N
Force to the left = 6.7N
Unknown:
Acceleration of the object = ?
Solution:
To solve this problem, we use newtons second law of motion;
Net force = mass x acceleration
Net force on object = Force to the right - Force to the left
Net force = 16.3N - 6.7N = 9.6N
So;
9.6 = 3.2 x a
a = 
= 3m/s²
