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3 years ago

How are radio telescopes

Chemistry

1 answer:

Alik [6]3 years ago

3 0

Keck telescope are more used in light, are visible, and is used in Hawaii.

For a regular radio telescope, it tracks and observes a much BIGGER waves, viewing in more detailed/more depth. Notice that light and it’s visibility isn’t much of a strong guidance!

Hope this helps!!

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Please someone help WHATS the answer

blsea [12.9K]

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2 years ago

cuantos gramos de oxígeno necesarios para que reaccionen completamente 28 gramos de etanol (c2h5oh) en la reaccion de combustion

tigry1 [53]

Answer:

I dont speak this language sorry :(

Explanation:

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3 years ago

As a storm from moves in, you notice that a column of mercury in a barometer rises to 736 mm. (a) Determine the air pressure. (b

sesenic [268]

Answer:

The air pressure is 9.8 *10^4 pa

The water will rise to a height of 10.0 meter

Explanation:

Step 1: Data given

As a storm from moves in, you notice that a column of mercury in a barometer rises to 736 mm.

Step 2: Calculate the air pressure

The Pressure against the mercury column = h*d*g = 0.736 * 13593 * 9.81 = 9.8 * 10^4 Pa

Step 3: Calculate the height of the water

Let the Pressure the water column for same pressure is h meter : -

9.8 * 10^4 = h*d*g

=>9.8*10^4 = h*1000*9.81

=>h = 10.0 meter

The water will rise to a height of 10.0 meter

5 0

3 years ago

The volume of a gas is always the same as the volume of __________.

Len [333]

The answer would be A: its container.

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7 0

3 years ago

Read 2 more answers

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers