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:A
Explanation: because The nucleus directs all of the cell's activities, including reproduction.
The
reaction is<span>
C</span>₂H₄ +
O₂ → CO₂ + H₂O<span>
To balance the equation, both side have same
number of elements. Here,</span>
In left hand
side has
in right hand side has
4 H atoms
2 H
atoms
2 C atoms 1 C
atom
<span>
2 O atoms 3 O
atoms
First, we have to balance number of C atoms and number of H atoms in both side.
To balance C atoms, '2' should be added before CO</span>₂ and to balance H atoms, '2' should be added
before H₂<span>O.
Then number of oxygen atoms is </span>2 x 2 + 2 = 6 in right hand side. So, 3 should be
added before O₂<span> in left hand side.
After balancing the equation should be,</span>
C₂H₄<span> + 3O</span>₂<span> → 2CO</span>₂<span> + 2H</span>₂O
<h2>C</h2>
Explanation:
The atomic number of S is 
So,number of electrons in S is
The electronic configuration of S is 
The orbital electronic configuration of S is 
So,the number of orbitals involved is 9.
Answer:
just use the tongs and put it on a plate
Explanation:
