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3 years ago

15

What would be another property that we Earthlings could use to classify stars?

1 answer:

Blizzard [7]3 years ago

8 0

Answer: At the point when space experts take a gander at an article's range, they can decide its arrangement dependent on these frequencies. The most well-known technique stargazers use to decide the sythesis of stars, planets, and different articles is spectroscopy. This spread-out light is known as a range.

Explanation:

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Plz help me it a timed test

Shtirlitz [24]

Answer:

B

Explanation:

6 0

3 years ago

A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

$\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%$

$1 - \frac{m(t+1)}{m(t)} = 0.002$

$1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002$

$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

$\tau \approx 499.500\,years$

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

The half-life of the radioactive isotope is 346 years.

6 0

3 years ago

Methane CH4 (g) reacts with oxygen gas to produce carbon dioxide and water. ____________________________________________________

nordsb [41]

<u>Answer:</u> The chemical equations are written below.

<u>Explanation:</u>

<u>For a:</u> Methane reacts with oxygen gas to produce carbon dioxide and water.

Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

<u>For b:</u> Butane reacts with oxygen gas to produce carbon dioxide and water.

This is also an example of combustion reaction.

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

<u>For c:</u> An aqueous solution of sulfuric acid reacts with aqueous potassium hydroxide to produce potassium sulfate and water.

When an acid reacts with a base, it leads to the formation of salt and water. This reaction is known as neutralization reaction

The chemical equation for the reaction of potassium hydroxide and sulfuric acid follows:

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

Hence, the chemical equations are written above.

4 0

3 years ago

Pls ans quicckk two q

Marina86 [1]

Answer:

there is nothing here

Explanation:

3 0

3 years ago

2.<br> A chemical property is a change in

liberstina [14]

Answer:

the chemical composition of a substance.

Explanation:

6 0

3 years ago