1. I am made up of two elements- one supports combustion, the other explodes with a spark. What am I? Organic or inorganic compound? 2. I am small, slim, hard and pointed. My partner hits me hard to get me in, and pulls me strongly to get me out. What am I? Metal or Non- metal

3. I am colorless liquid. I can kill germs upon contact. I evaporate fast when applied to your skin. What am I? Solution or Suspension?

4. I am usually slim and thin. My skin is smooth. My head burns to give off light. What am I? Substance or Mixture?

5. I am a white liquid. Babies like to drink me. Your mother also tells you to drink me every day. What am I? Substance or Mixture?

6. I am yellow and shiny. Because I do not react with air or acid, I am used for jewelry. What am I? Element or Compound?

2. I am soft and yellow. I make your hot pandesal tasty. What am I? Substance or Mixture?

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Do not include the unexpected results

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that  0.249 or 0.25  grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.