Answer:
Above the Curie temperature, a magnet permanently loses all or some of its magnetism. External magnetic fields: Strong, opposing magnetic fields can cause the magnetic domains to lose their orientation and relax into a lower state of energy where they are not aligned.
Explanation:
Answer:
Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.
Explanation:
Molarity of the solution is the moles of compound in 1 Liter solutions.

Mass of NaOH = x
Molar mass of NaOH = 40 g/mol
Volume of the NaOH solution = 450 mL =- 0.450 L ( 1 ml = 0.450 L)
Molarity of the solution of NaOH = 0.250 M


Solving for x:
x = 4.5 g
Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.
Answer:
The heat of combustion is -25 kJ/g = -2700 kJ/mol.
Explanation:
According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.
Qcomb + Qcal = 0
Qcomb = - Qcal
The heat absorbed by the calorimeter can be calculated with the following expression.
Qcal = C × ΔT
where,
C is the heat capacity of the calorimeter
ΔT is the change in temperature
Then,
Qcomb = - Qcal
Qcomb = - C × ΔT
Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ
Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:

The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:

Answer: Cs because its further to the right on the table, meaning its more reactive.
Explanation:
The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
= 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
= 0.055/ ( 355 ×10^-3)
= 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)
1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
= 0.0563 moles
[HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
= 0.0563/(355×10^-3)
= 0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation
Ka of H2PO4- = 6.20 × 10^-8
[H+] =Ka*([H2PO4-]/[HPO4(2-)]
= (6.20 ×10^-8)×(0.155/0.1586)
= 6.059 ×10^-8 M
pH = - log[H+]
= - log (6.059×10^-8)
= 7.218