Answer:
At the equivalence point of the titration of a monoprotic weak acid with a strong base: <em>B. the moles of strong base added must equal the moles of weak acid.</em>
Explanation:
In every titration, the equivalence point is defined as the point where the moles of the titrant and analyte are equal. For every acid-base titration, the equivalence point is defined as the point where the moles of the base is equal to the moles of the acid.
If the solutions of the acid and base are at a different concentration the volume added from the buret will not be the same as the volume of the analyte.
Hey there!
Molar mass CuSO4 = <span>159.609 g/mol
</span><span>Note that the molarity is in the enunciation of the issue, then the problem asks the mass of the solute :
Volume in liters of solution:
150 mL / 1000 => 0.15 L
</span><span>Calculation of quantity in Moles of the solute :
</span>
1 L ----------------- 0.300 M
0.15 L ------------ ( moles )
Moles CuSO4 = 0.15 * 0.300 / 1
Moles CuSO4 = 0.045 moles
Therefore :
1 mole CuSO4 --------------------- 159.609 g
0.045 moles CuSO4-------------- mass
mass = 0.045 * 159.609 / 1
mass = 7.182405 g of CuSO4
As a refrigerator lowers the temperature of the items it contains, the thermal energy that is removed from these items is <span>released to the surroundings. Thermal energy isn't stored in a fridge so that causes the temperature to increase.
Answer: Letter B </span><span>✅
</span><span>
Hope that helps. -UF aka Nadia</span>
Answer:
The volume occupied is 25.7 L
Explanation:
Let's replace all the data in the formula
P . V = N . k . T
N = nNA
1 gm.mole . 6.02x10²³
k = Boltzmann's contant
T = T° in K
1 atm = 1.013 × 10⁵ N/m2
1.013 × 10⁵ N/m2 . Volume = 6.02x10²³ . 1.38065 × 10⁻²³ N · m/K . 313.7K
Volume = (6.02x10²³ . 1.38065 × 10⁻²³ N · m/K . 313.7K) / 1.013 × 10⁵ m2/N
Volume = 2607.32 N.m / 1.013 × 10⁵ m2/N = 0.0257 m³
1 dm³ = 1 L
1m³ = 1000 dm³
25.7 L