Use the following equation to answer the questions below. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

1. What type of reaction is this? Why do you say it is this type of reaction? Support your claim with evidence. (2 points, 1 point valid type of reaction, 1 point valid evidence)

2. What is the mole ratio of methane to water in the reaction? (1 point correct mole ratio)

3. How many moles of water are produced from the reaction of 50.0g of methane (CH4)? Show your work (2 points, 1 point valid answer, 1 point valid work)

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts

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.