Solids are compounds whose atomic bonds are rigid, which doesn't let the atoms move around freely
Answer:
Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.
Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.
Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.
Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.
Re-weigh the crucible and contents once cold.
Calculation:
Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)
Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment
Calculate the number of moles of anhydrous copper(II) sulfate formed
Calculate the number of moles of water driven off
Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed
Write down the formula for hydrated copper(II) sulfate.
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Explanation:
The circulatory system picks up nitrogenous wastes from the cells and delivers them to the kidneys. The kidneys remove these wastes from the blood and concentrates them into the urine that is eliminated from the body.
Answer:
B = (2.953 × 10⁻⁹⁵) N.m⁹
Explanation:
At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.
That is,
Fa + Fr = 0
Fa = - Fr
Fa = (|q₁q₂|)/(4πε₀r²)
Fr = -B/(r^n) but n = 9
Fr = -B/r⁹
(|q₁q₂|)/(4πε₀r²) = (B/r⁹)
|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C
(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²
r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m
(k|q₁q₂|)/(r²) = (B/r⁹)
(k × |q₁q₂|) = (B/r⁷)
B = (k × |q₁q₂| × r⁷)
B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]
B = (2.953 × 10⁻⁹⁵) N.m⁹
