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2 years ago

How are NGC 1427A and U different? How are they the same?

Chemistry

2 answers:

irga5000 [103]2 years ago

5 0

Answer:

Sample Response: NGC 1427A has no general shape, so it is an irregular galaxy. U has a bulge in the center and arms, so it is a spiral galaxy. They are similar in that both contain plenty of dust and gas. Both also have active star-forming sites.

Explanation:

Which ideas did you include in your answer? Check all that apply.

✅ NGC 1427A is an irregular galaxy.

✅ U is a spiral galaxy.

✅ Both galaxies contain plenty of dust and gas.

✅ Both galaxies have active star-forming sites.

Ik I'm late but YW

agasfer [191]2 years ago

3 0

Answer:

that looks pretty and also well NGC 1427A has no general shape, so it is an irregular galaxy. U has a bulge in the center and arms, so it is a spiral galaxy. They are similar in the both certain plenty of dust and gas. Both also have active star-forming sites.

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A sample of gas is placed into an enclosed cylinder and fitted with a movable piston. Calculate the work (in joules) done by the

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2 years ago

could you please explain operations with scientific notation using this example. 5 × 10³ + 4.3 × 10⁴ i dont understand how to so

TEA [102]

Explanation:

The scientific notation:

$a\times10^k$

where

$1\leq a$ and k is integer.

We have the example:

$(5\times10^3)+(4.3\times10^4)$

You can write the numbers in a "normal" form:

$5\times10^3=5\times1000=5000\\\\4.3\times10^4=4.3\times10000=43000$

Make the sum:

$5000+43000=48000$

And next write it in the scientific notation:

$48000=4\underbrace{8000}_{\leftarrow4}=4.8000\times10000=4.8\times10^4$

<h3>Other method:</h3>

You can add numbers in scientific notation if the power of tens in both number is the same.

Therefore you must convert the first number:

$5\times10^3=0.5\times10\times10^3=0.5\times10^4$

Now, you can make the sum:

$(5\times10^3)+(4.3\times10^4)=0.5\times10^4+4.3\times10^4=(0.5+4.3)\times10^4=4.8\times10^4$

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3 years ago

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(

likoan [24]

Balanced Half reactions are:

At anode 2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$

At Cathode: 2 $H^{+}$ + $2e^{-}$ ==> H₂

Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, $H^{+}$ , $Mn^{2+}$ and $Cl^{-}$

Now at Anode reaction will occur as given:

2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$ (will occur)

At Cathode:

2 $H^{+}$ + $2e^{-}$ ==> H₂ (will occur)

At Cathode:

$Mn^{2+}$ + $2e^{-}$ ==> Mn (This reaction will not occur)

The deposition of solid Mn will not occur because in aqueous solution, $H^{+}$ will be reduced before $Mn^{2+}$ .

The reduction potentials for $H^{+}$ is zero whereas reduction potential for $Mn^{2+}$ is - 1.18V.

The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.

To learn more about the half reaction please click on the link brainly.com/question/13186640

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1 year ago