Answer:
5.3 × 10⁻¹⁷ mol·L⁻¹
Explanation:
Let <em>s</em> = the molar solubility.
Cu₂S(s) ⇌ 2Cu⁺(aq) + S²⁻(aq); K_{sp} = 6.1 × 10⁻⁴⁹
E/mol·L⁻¹: 2<em>s</em> <em>s
</em>
K_{sp} =[Cu⁺]²[S²⁻] = (2<em>s</em>)²×<em>s</em> = 4s^3 = 6.1 × 10⁻⁴⁹
![s = \sqrt[3]{1.52 \times 10^{-49}} \text{ mol/L} = 5.3 \times 10^{-17} \text{ mol/L}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%5B3%5D%7B1.52%20%5Ctimes%2010%5E%7B-49%7D%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%205.3%20%5Ctimes%2010%5E%7B-17%7D%20%5Ctext%7B%20mol%2FL%7D)
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Answer:
Consequently, what happens when gas obtained by heating slaked lime and ammonium chloride is passed through copper sulphate solution? The HCl in the gas mixture will form hydrochloric and the H+ will react with some of the NH3(aq), forming NH4^+, and with some of the SO4^2-, forming HSO4^-.
Answer:
<h3>The man's average body temp. will fall by 0.6°C to 38.4°C</h3>
Explanation:
The enthalpy (heat content) of the water, using a datum of 0°C, is
Hw = Mw kg x Cp,w (specific heat capacity kJ/kg °C) x Tw °C
= 1 kg x 4.18 kJ/kg°C x 3°C = 12.5 kJ
Hman (pre drink) = 68 kg x 3.6 kJ/kg/°C x 39°C = 9547 kJ
So Hman (post drink) = H (pre drink) + Hw = 9547 + 12.5 = 9559.5 kJ because no heat is lost immediately.
But Hman (after drink) has mass 68 + 1 = 69 kg
Also his new Cp will be approx (3.6 x 68/69) + (4.18 x 1/69) = 3.608 kJ/kg°C
So Hman = 9559.5 kJ (from above) = 69 kg x 3.608 kJ/kg°C x Tnew
Therefore the man's new overall temp. = 38.4°C which is a drop of 0.6°C
2. Left to right (smallest to largest) F-, Mg2+, Na+, and K+
5. Cs, Ga, As, N, O
6. Cs or H (probably cesium)
7. In or In 2+ (not sure which)
Hope this helps, I've never really seen questions like this before.
