Chemistry, 18.03.2020 23:26 dyhji

1. a person walks 3 kilometers in an hour, rests for an hour, and then walks 3 more kilometers in another hour. What is the average speed of the person over the three hours?

2. a car travels north for 20 km. The car then travels south for 35 km. If north is taken as the positive direction, what is the displacement of the car?

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Answer from: yolo123321

Hey there!

1. To find the average speed of something, you use the formula d=rt, or distance equals rate (speed) multiplied by time. We can plug what we have into this equation to find the average speed. The distance is 6 km, and the time is three hours.

6=3r

Now we sole

r=2

Therefore, the average speed is 2 km per hour.

2. If we were to put this on a vertical number line, the car would be at -15. Displacement is the change in original position. If it was originally at point 0, it's displacement is 15 km south. Because it says that north is a positive direction, that must mean the south is negative.

Therefore, your answer is -15.

I hope this helps! Have a great day!

Answer from: betoo9090

1. 2km/h 2. -15km

Explanation:

1. speed=whole distance/time=(3+0+3)/3=2

2. 20-35=-15

Answer from: Quest

answer: they produce more slow than unicellular organisms.

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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