Chemistry, 07.05.2021 19:30, isaaccott013

Convert 801 grams of aluminum oxide (AI2O3) to moles. Start by finding the molar mass of AI2O3 Report your answer to 3 significant figures

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answered: Guest

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

$^{222}_{86} rn+ ^{4}_{2} he$ →

answered: Guest

Mass is the amount of matter in an object. weight is the measurement of the pull of gravity on an object. say for example, you travel to the moon which has a different gravitational field compared to the gravitational field of earth. (gravitational fields: earth = 1, moon = 0.165) now when you are on the moon, the pull of gravity on you is different than on the earth. and as weight is the measurement of the pull of gravity on an object, that means yes, your weight can and will change on the moon. however, on the moon, on earth or even on jupiter, the amount of mass inside of you remains the same. meaning your physical mass remains constant. in conclusion it is possible for your weight to change as your mass remains constant.

answered: Guest

Bthats what im doing to and i figured it out it is b

answered: Guest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ 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: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.