What Is A Nebula?

2 answers:

7 0

a nebula is a cloud of gas and dust in outer space, visible in the night sky either as an indistinct bright patch or as a dark silhouette against other luminous matter.

therfore your answer would be C cloud of dust and gas in space

question answered by

(jacemorris04)

ludmilkaskok [199]4 years ago

6 0

The answer to your question is,

C. Cloud of dust and gas in space.

-Mabel <3

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The point in a titration at which the indicator changes is called the

irakobra [83]

Answer:

End point

Explanation:

The point at which the indicator changes color is called the endpoint. So the addition of an indicator to the analyte solution helps us to visually spot the equivalence point in an acid-base titration

#correct me if I'm wrong

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3 years ago

Which is warmer -30C or -30 F?

Burka [1]

-30°C is more warmer than -30°F...!

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Which of these elements exhibits chemical behavior similar to that of oxygen? select one:

scoray [572]

Answer:
Sulfur (Option-C) exhibits chemical behavior similar to that of oxygen.

Explanation:
Sulfur has same chemical properties as that of Oxygen because both of them belongs to same group in the periodic table. Also, the similarity of chemical behaviour among the group members is due to same number of electrons in their valence shells.
For examole, the electronic configuration of Oxygen is,

1s², 2s², 2p⁴

There are six valence electrons in the valence shell (i.e. 2) of Oxygen.

Now for Sulfur,
1s², 2s², 2p⁶, 3s², 3p⁴

There are six valence electrons in the valence shell (i.e. 3) of Sulfur.

Therefore, both elements tends to gain 2 electrons in a reaction and form O⁻² and S⁻² respectively.

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4 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$