<span>The process of heating water above its boiling point without it boiling and it remaining liquid is called superheating. It is heated under pressure and can be heated to temperatures of 374 degrees Celsius.</span>
2NaCl + Br2
This is because Na (Sodium) still needs to retain the coefficient or amount of two. Since Chlorine already has 2 on the reactants side, having a two in front would make sense so both elements can have a two.
Bromine would have the subscript of two as in the reactants side of the equation, it is also under the coefficient of two. Thus, it would need to carry it to the products side in order to keep the equation balanced.
Hope this helps!
Answer:
See explanation
Explanation:
Dilution law= C1V1=C2V2.
Where C is the concentration in mol per dm cube and V= volume in dm cube.
From the questions the parameters given are; V= 42.5ZmL, concentration or molarity of HCl= 0.193M and the molarity of KOH = 0.125M.
(a). At 0mL where the base KOH has not been added yet.
Moles of HCl = molarity × volume.
Moles of HCl = 0.193 × 42.5/1000
Moles of HCl= 0.00820205
pH=- log 0.0082025
pH= 2.086
Approximately, pH=2.1.
(b). At 3mL of 0.125M KOH
3/1000 × 0.125 mol/L
= 0.003L× 0.125 M
= 0.000375 moles KOH
Determining the number of moles of HCl, we have;
42.5/1000 × 0.193 mol/L
= 0.0082025
Final volume= 3mL+42.5mL= 45.5 ml
Final[HCl] = 0.0082025 - 0.000375 = 0.0078275.
Therefore, 0.0078275/ 0.0455
= 0.1720
pH= -log 0.1720
pH= 0.76.
(c). 5mL of 0.125M KOH
5/1000 × 0.125
= 0.005×0.125
= 0.000625 mol KOH
42.5/1000 × 0.193 mol/L
= 0.0082025 mol
Final volume= 5mL + 42.5= 47.5 mL.
0.0082025-0.000625= 0.0076
Final [HCl] = 0.0076/0.0475
= 0.16
pH= -log 0.16
pH= 0.79.
Answer:
Explanation:
Hello there!
In this case, according to the given problem, it is possible to use the given molar masses of NaCl, KCl and LiCl in order to calculate the total moles, yet we calculate the moles of each salt first as shown below:
Then, we add them all together to obtain:
Regards!
A temperature change maybe?
