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nadya68 [22]
3 years ago
11

True or False Helium belongs to Noble Metals

Chemistry
2 answers:
Lostsunrise [7]3 years ago
6 0

Answer:

This answer is "True"

Vlada [557]3 years ago
6 0
The answer would be *true*
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If you have 3.0 moles of argon gas at STP, how much volume will the argon take up?
Anestetic [448]
If you have 3.0 moles of argon gas at STP u would take up 2.5 volume
3 0
3 years ago
Which processes of the water cycle involve the movement of water in the liquid state?
dangina [55]

Answer:

Surface runoff and condensation

Explanation:

Let's define each of the given processes in order to understand them better:

  • evaporation is a process in which liquid phase transforms into a gas phase;
  • precipitation is a process in which we produce a solid phase, usually this is the case when we precipitate a salt out of a solution, analogy of precipitation for water would be transformation from a liquid to a solid phase, such as freezing;
  • surface runoff is a process in which water flows over the surface of a land without any change in its phase;
  • condensation is a process in which a gas transforms into a liquid.

All in all, notice that surface runoff keeps water in its liquid state, while all the other three options consider phase change. The only phase change of interest is condensation: we produce liquid water from water vapor and then we can analyze its movement in the liquid state.

6 0
3 years ago
Will bromine react with sodium and why?
Archy [21]

<span><span>When you write down the electronic configuration of bromine and sodium, you get this

Na:
Br: </span></span>

<span><span />So here we the know the valence electrons for each;</span>

<span><span>Na:  (2e)
Br:  (7e, you don't count for the d orbitals)

Then, once you know this, you can deduce how many bonds each can do and you discover that bromine can do one bond since he has one electron missing in his p orbital, but that weirdly, since the s orbital of sodium is full and thus, should not make any bond.

However, it is possible for sodium to come in an excited state in wich he will have sent one of its electrons on an higher shell to have this valence configuration:</span></span>

<span><span /></span><span><span>

</span>where here now it has two lonely valence electrons, one on the s and the other on the p, so that it can do a total of two bonds.</span><span>That's why bromine and sodium can form </span>

<span>
</span>

4 0
3 years ago
A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0
3 years ago
Binary ionic compound of calcium and oxygen
Vlad1618 [11]
Socratic helps for you page
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