Chemistry, 12.10.2020 20:01 LuluMathLover101

A student collected the data shown in the table below during an experiment. Liquid Characteristics
Liquid Characteristics
Boiling Point Freezing Point Color
Toxicity
Mercury 357 °C -39 °C
shiny silver gray hazardous if tube is broken
Alcohol 78 °C
- 115 °C colorless relatively non-hazardous
Based on the data, which of the following conclusions can be made about the use of mercury and alcohol thermometers? (5 points)
An alcohol thermometer can easily measure the boiling point of a colorless liquid that boils at 100 °C.
A mercury thermometer can easily measure the boiling point of a colorless liquid that boils at 82 °C.
A mercury thermometer can easily measure the freezing point of a colorless liquid that freezes at -50 °C.
An alcohol thermometer can easily measure the freezing point of a colorless liquid that freezes at -110 °C.

A student collected the data shown in the table below during an experiment.

Liquid Characteristic

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it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

Answer from: Quest

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.

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Electrons are emitted

Answer from: Quest

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.

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A student collected the data shown in the table below during an experiment. Liquid Characteristics<...

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