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3 years ago

How are scientists able to study the composion of a star that is located out of the solar system

1 answer:

5 0

The most common method astronomers use to determine the composition of stars, planets, and other objects is spectroscopy. Today, this process uses instruments with a grating that spreads out the light from an object by wavelength. This spread-out light is called a spectrum.
I hope this helps.

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Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago

Which of the following solutions would have [Fe3+] = 0.020 M?

Shalnov [3]

Answer:

None of the given options

Explanation:

Let's go case by case:

A. No matter the volume, the concentration of Fe(NO₃)₃ (and thus of [Fe³⁺] as well) is 0.050 M.

B. We can calculate the moles of Fe₂(SO₄)₃:

0.020 M * 0.80 L = 0.016 mol Fe₂(SO₄)₃

Given that there are two Fe⁺³ moles per Fe₂(SO₄)₃ mol, in the solution we have 0.032 moles of Fe⁺³. With that information in mind we <u>can calculate [Fe⁺³]</u>:

0.032 mol Fe⁺³ / 0.80 L = 0.040 M

C. Analog to case A., the molar concentration of Fe⁺³ is 0.040 M.

D. Similar to cases A and C., [Fe⁺³] = 0.010 M.

Thus none of the given options would have [Fe⁺³] = 0.020 M.

7 0

2 years ago

How are the orbits of asteroids similar to, or different from planetary orbits?

Strike441 [17]

Aside from banks, what are 2 other places you might be able to borrow
money? And can you help with more questions on my page thank you so much I really appreciate you and everything you do for this app

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2 years ago

NEED HELP NOW 62 POINTS WILL MARK BRAINLIEST!!! PLZ ANSWER ALL AND DO NOT COPY OFF WEBSITE QUESTIONS DOWN BELOW

Karolina [17]

#19 due to weathering and erosion the rock from the stream was smoothened and without as much the rock from deep underground was/is very rugged and natural.

#18 Due to wind patterns the sand dunes are blew around and the sands shift and fly around with nothing to keep the sand in place the sand can go long distances or come long distances.

4 0

2 years ago

Determine the pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C. Carry out two c

Nuetrik [128]

The pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C is 101.94atm.

<h3>How to calculate pressure?</h3>

The pressure of an ideal gas can be calculated using the following formula:

PV = nRT

Where;

P = pressure
V = volume
n = number of moles
R = gas law constant
T = temperature

According to information in this question;

T = 25°C = 25 + 273 = 298K
V = 244.6mL = 0.24L
R = 0.0821 Latm/Kmol

P × 0.24 = 1 × 0.0821 × 298

0.24P = 24.47

P = 24.47/0.24

P = 101.94atm

Therefore, the pressure in atm exerted by 1 mole of methane placed into a bulb with a volume of 244.6 mL at 25°C is 101.94atm.

Learn more about pressure at: brainly.com/question/11464844

3 0

2 years ago