Base on your question where a concentration cell consist of two SN/SN2+half cells. The solution in one half cell A is 0.13M SN(NO3)2 and is 0.87 M Sn(NO3)2 in the other half cell to get the cell potential at 25 degree the answer is 0.059/2 log0.13/0.87
Answer:
<h2>4.55 L</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we're finding the new volume
We have
We have the final answer as
<h3>4.55 L</h3>
Hope this helps you
First a balanced reaction equation must be established:
→
Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[
4Al +
3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
=
3.63 mol (to 2 sig. fig.)
Thus it can be concluded that
3.63 moles of oxygen is needed to react completely with 145 g of aluminum.
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
To learn more about stoichiometric quantities visit:
<h3>
brainly.com/question/28174111</h3>
#SPJ4
Answer: Do want the experiment done or help of the experiment