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3 years ago

Current divides between the available path?

1 answer:

3 0

The current divides according to the resistance; more current in the lower resistance, less in the higher resistance. This is called “parallel branches” or paths. The voltage is the same for both (all) branches.

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What elements are in nickel oxide

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nickel oxide is the chemical compound with a formula called NiO.

what are the elements in nickel oxide?

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4 years ago

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What makes gabbro a rock and not a mineral

lozanna [386]

Gabbro is usually made up of such as pyroxene, plagioclase, amphibole and olivine which makes it an intrusive igneous rock. It is not considered as a mineral since it is composed of an aggregate of minerals or non-minerals and does not have one specific chemical composition.

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3 years ago

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How many moles of O2 are needed to produce 3.95 moles of Fe2O3

Minchanka [31]

Answer:

4Fe+3O2- 2Fe2O3

the ratio: 3:2

x:3.95

x=5.92 moles of O2

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3 years ago

Even before modern observations provided evidence supporting the theory of plate tectonics, people developed theories that the c

Deffense [45]

Answer:

people or scientist tend to observe the nature of the continents that described by jigsaw fit,geological similarity among the continents

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3 years ago

a 182.4g sample of germanium-66 is left undisturbed for 22.5 hours. At the end of that period, only 5.70g remain. What is the ha

Nana76 [90]

Answer:

Approximately $4.5\; \text{hours}$ .

Explanation:

Calculate the ratio between the mass of this sample after $22.5\; \text{hours}$ and the initial mass:

$\displaystyle \frac{5.70\; \rm g}{182.4\; \rm g} \approx 0.03125$ .

Let $n$ denote the number of half-lives in that $22.5\; \text{hours}$ (where $n\!$ might not necessarily be an integer.) The mass of the sample is supposed to become $(1/2)$ the previous quantity after each half-life. Therefore, if the initial mass of the sample is $1\; \rm g$ (for example,) the mass of the sample after $\! n$ half-lives would be ${(1/2)}^{n}\; \rm g$ . Regardless of the initial mass, the ratio between the mass of the sample after $n\!\!$ half-lives and the initial mass should be ${(1/2)}^{n}$ .

For this question:

${(1/2)}^{n}} = 0.03125$ .

Take the natural logarithm of both sides of this equation to solve for $n$ :

$\ln \left[{(1/2)}^{n}}\right] = \ln (0.03125)$ .

$n\, [\ln(1/2)] = \ln (0.03125)$ .

$\displaystyle n = \frac{\ln(0.03125)}{\ln(1/2)} \approx 5$ .

In other words, there are $5$ half-lives of this sample in $22.5\; \text{hours}$ . If the length of each half-life is constant, that length should be $(1/5) \times 22.5\; \text{hours} = 4.5\; \text{hours}$ .

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3 years ago