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Luba_88 [7]
4 years ago
5

Which molecules are pure elements?

Chemistry
2 answers:
Elenna [48]4 years ago
8 0
Well none since molecules are a group of two or more atoms electrically bonded with one another. However, there are gases that does not naturally bond due to their stability and can be found in nature as pure elements. But these are not considered as molecules.

(By the way, these gases are the noble gases that can be found on the last column of the periodic table) 
Vaselesa [24]4 years ago
4 0

Answer: the molecules formed by one kind of atom are pure elements


Explanation:


1) The definition of molecules is the covalent bonding of two or more atoms.


2) Under that definition you must have at least two atoms to be considered a molecule.


3) Pure elements are the 118 elements that you find in the periodic table: each contain only one kind of atom.


4) Then, molecules of pure elements, are two or more atoms of the same kind linked by covalent bonds.


5) With that these some examples of molecules that are pure elements:


i) O₂

ii) N₂

iii) F₂

iv) Cl₂

v) Br₂

vi) O₃


6) And here some examples of molecules that are not pure elements (there are many, many):


i) CH₄

ii) CH₃CH₂CH₃

ii) CH3CH₂CH₂CH₂CH₂CH₂CH₂CH₃


7) Note that ionic compounds are excluded of the definition of molecules, since they are not formed by covalent bonds but by ionic bonds. For example, NaCl is not a molecule.

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​If you needed a 1.5 x 1 0-4 M solution of a compound that has a molar mass of 760 g/mol, what would it concentration be in part
motikmotik

Answer:

114 ppm

Explanation:

Data obtained from the question include:

Conc. of compound in mol/L = 1.5×10¯⁴ mol/L

Molar mass of compound = 760 g/mol

Conc. in ppm =..?

Next, we shall determine the concentration of the compound in grams per litre (g/L) . This is illustrated below:

Conc. in mol/L = conc. in g/L / Molar mass

1.5×10¯⁴ = conc. In g/L / 760

Cross multiply

Conc. in g/L = 1.5×10¯⁴ x 760

Conc. in g/L = 0.114 g/L

Next, we shall convert 0.114 g/L to milligrams per litre (mg/L). This is illustrated below:

1 g/L = 1000 mg/L

Therefore, 0.114 g/L = 0.114 x 1000 = 114 mg/L

Finally, we shall convert 114 mg/L to parts per million (ppm). This is illustrated below:

1 mg/L = 1 ppm

Therefore, 114 mg/L = 114 ppm

From the calculations made above,

1.5×10¯⁴ mol/L Is equivalent to 114 ppm.

6 0
3 years ago
A saturated solution of potassium chlorate is dissolved in 100g of water. If the saturated solution is cooled from 90 degrees Ce
dedylja [7]

Mass would crystallize out : 24 g

<h3>Further explanation </h3>

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility

  • 1. Temperature:
  • 2. Surface area:
  • 3. Solvent type:
  • 4. Stirring process:

We can use solubility graph to solve this.

solubility KClO₃ at 90°C=52 g

solubilty KCl₃ at 60°C= 28 g

Mass would crystallize out = 52 - 28 =24 g

4 0
3 years ago
Why can carbon bond to many elements simultaneously? because its +4 valence forms ionic bonds because it has a natural ionic cha
mina [271]
<span>Because it is capable of forming multiple covalent bonds. I hope 15 hours isn't too late to answer your question.

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6 0
4 years ago
Read 2 more answers
What is the destiny of an object that has a mass of 20 grams and a volume of 10 millimeters
blondinia [14]

Answer:

2gcm⁻³

Explanation:

Given parameters:

Mass of the object  = 20g

Volume = 10mL

Unknown:

Density of the object  = ?

Solution:

Density of a body is its mass per unit volume;

   Density  = \frac{mass}{volume}  

We need to convert mL to cm³ for the volume

     1mL  = 1cm³;

        10mL is therefore, 10cm³  

So;

       Density  = \frac{20}{10}   = 2gcm⁻³

7 0
3 years ago
Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate
nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

N=N_A\times n

Charge on N electrons : Q

Q = N\times 1.602\times 10^{-19} C

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A

18.0 Ampere is the size of electric current that must flow.

3 0
3 years ago
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