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Novay_Z [31]
3 years ago
9

Write a balanced equation for the reaction between aqueous strontium chloride and aqueous lithium phosphate to form solid stront

ium phosphate and aqueous lithium chloride.
Chemistry
1 answer:
Marat540 [252]3 years ago
5 0
First, let's write the chemical formula for each of the substances mentioned in the problem.

Strontium Chloride: SrCl₂
Lithium Phosphate: Li₃PO₄
Strontium Phosphate: Sr₃(PO₄)₂
Lithium Chloride: LiCl

So, the balanced chemical reaction is:

<em>3 SrCl₂ (aq) + 2 Li₃PO₄ (aq) ---> Sr₃(PO₄)₂ (s) + 6 LiCl (aq)</em>

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What is the name of the molecule shown below?
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The answer is B, Ethanal
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Van der Waals forces hold molecules together by: A. moving electrons from one molecule to another. B. attracting a lone pair of
Rasek [7]

The forces that are existing between molecules are known as intermolecular forces. These forces which are weaker than ionic and covalent bonds are classified into three types'

--> dipole-dipole attraction

--> VAN DER WAALS FORCES and

--> hydrogen bonding.

Van Der Waals forces was postulated by a Dutch physicist known as Van Der Waals. He postulated the existence of weak, short-range forces of attraction, which are independent of normal bonding forces, between non-polar molecules. He came to this conclusion after studying the behaviour of real gases at low temperatures and high pressures that:

--> electrons in a non-polar molecule such as hydrogen are close to one nucleus as to the other, although momentary concentration at one end of the molecule may occur,

--> this momentary concentration of electron cloud on one side create a temporary dipole in the hydrogen molecule, that is, one side of the molecule acquires a partial negative charge while the other side acquires a partial positive charge of equal magnitude,

--> the temporary dipole induces a similar dipole in an adjacent molecule,

--> this results in a temporary dipole-induced dipole attraction between the positive and negative ends of the adjacent molecules.

This is how weak Van Der Waals forces are set up. Therefore, option C is CORRECT which states that VAN DER WAALS forces hold molecules together by inducing temporary dipoles that attract each other.

Learn more about Van Der Waals forces here:

brainly.com/question/11457190

6 0
3 years ago
PLEASE ANSWER...........WILL MARK BRAINLIEST!!!!!!!
fomenos

Explanation:

Alkali metal cations reacts vigorously with halogens to form ionic/electrovalent compounds.

  • The alkali metals are about the most reactive metals and they belong to group I .
  • In their outermost shell, they have one valence shell electron.
  • Alkali metals are largely electropositive willing to release their valence electron to attain stability.
  • The halogens have seven electrons in their outermost shell. To complete their configuration, they need just one electron.
  • All the halogens are strongly electronegative and they are readily reactive.
  • In the vicinity of group 1 elements, they react vigorously.
  • This is due to the electrostatic attraction between the metal and non-metal ion formed.
  • Therefore, they form strong ionic bonds between their compounds.

Learn more:

sodium alkali and halogens brainly.com/question/6324347

#learnwithBrainly

4 0
3 years ago
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Which reaction best represents the complete combustion of ethane?
baherus [9]
Answer:

Combustion reaction

2C2H6+7O2 → 4O2+6H2O

Explanation:

In a combustion reaction with a hydrocarbon in the reactant side you will always have O2 as another reactant. As you will always have CO2 and H2O as the products.

Knowing that much you can set up your reaction equation..

C2H6+O2→ CO2+H2O

Now the balancing can begin. Balancing hydrocarbon combustion reactions can be tricky, but if with practice they can be really fun and very rewarding.

Start with the C atoms first and move to the H atoms next. It's easier to leave the O2 to the last, it has a way to alter the equation.

Initially, you would arrive at this, before the O2 has been balanced:

C2H6+O2→2CO2+3H2O

But, as you can see, you have an odd amount of O2 on the product side. In this case, you have to find the common factor of the amount of O on the product side and 2, Because of the O2 diatom. Therefore, 14 would be the lowest common factor of 2 and 7.

I hope it helped you!
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