We are given the density of mercury which is equal to 13.6 g/ml. Substances which sink in this metal should have densities higher than the density of the metal, otherwise they would float. The substance thta would sink in mercury is C. gold
Answer:
Saturated solution.
Explanation:
- Kindly see the attached image, the curve with brown line is for KClO₃, and according to the curve at 30
°C the solubility of KClO₃ per 100 g of water is 10 g.
So, 10.0 g of KClO₃ forms a saturated solution.
Answer:
V NH3 = 304.334 L
Explanation:
- 1 mol ≡ 6.02 E23 molecules
⇒ moles NH3 = (7.50 E24 molecules)×(mol/6.022 E23 molecules)
⇒ mole NH3 = 12.454 mol
assuming ideal gas:
STP:
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
⇒ V NH3 = RTn/P
⇒ V NH3 = ((0.082 atm.L/K.mol)×(298 K)×(12.454 mol))/(1 atm)
⇒ V NH3 = 304.334 L
<span>By definition:
pH = pKa + log [acetate]/ [acetic acid]
so
5.02 = 4.74 + log [acetate] / 10 mmole
10mmole = 10/1000 = 0.01 mole
5.02 = 4.74 + log [acetate] / 0.01
5.02 - 4.74 = 0.28 = log [acetate] /0.01
10^0.28 = </span><span>1.90546</span> = [acetate] / 0.01 <span>
[acetate] = 0.019 mole
= 19 millimoles
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