What is the ratio Al 3+ ions to S 2- ions in a neutral compound

What happens if a solid changes the state of substance to liquid?

kati45 [8]

It would had to have melted in order to change the properties from a solid into a liquid. It also depends on the type of solid wheither a chemical or a ice cube. If it was a ice cube if would dilutte the liquid solution and changes the taste ot texture (feel) if it was a chemical the properties could change the liquid into a solid only depending on th type of chemical. EX..Liquid Nitrogen and a piece of raw chiken the outcome after putting the chicken in the nitrogen would be frozen. Because the chemical Nitrogen freezes the particles which makes the chicken frozen.

4 0

3 years ago

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Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Sauron [17]

Answer:

The concentration of COF₂ at equilibrium is 0.296 M.

Explanation:

To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the <em>concentrations</em> or <em>changes in concentration</em> in that stage. For this reaction:

2 COF₂(g) ⇌ CO₂(g) + CF₄(g)

I 2.00 0 0

C -2x +x +x

E 2.00 - 2x x x

Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".

$Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852$

The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M

6 0

3 years ago

OMG I GOT A 50% ON MY QUIZ BUT ALL I DID WAS ASK Y'ALL. I NEED COMFORTAGE AND DO NOTTTT FORGET I AM AUTISTIC GUYS!!! THANKX

Maslowich

Answer:just breath its ok just use brainly

Explanation:

7 0

3 years ago

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Please help with this question

mixas84 [53]

Answer:this Covalent bond

Explanation:Covalent bond involves a pair of electrons being shared between atoms.

Hope this helps!

8 0

3 years ago

Calculate the volume of 1.0 x 10-4 M CV+ solution that needs to be added to a 25.0 mL volumetric flask and diluted with deionize

xenn [34]

Answer:

The answers are: 2.5 ml (first part) and 3.0 ml (second part)

Explanation:

In order to calculate volumes required to prepare diluted solutions we use the following equation:

Vc x Cc = Vd x Cd

Where Vc and Cc are the volume and concentration respectively of concentrated solution (higher concentration) whereas Vd and Cd are volume and concentration of diluted solution (lower concentration).

In both problems we want to prepare a diluted solution and we know the final concentration (Cd) and final volume (Vd) and the initial concentration (Cc).

In first part, we have: Cc= 1.0 10⁻⁴ M; Vd= 25 ml; Cd= 1.0 10⁻⁵ M

Vc= Vd x Cd / Cc= (25 ml x 1.0 10⁻⁵ M)/1.0 10⁻⁴ M = 2.5 ml

Notice that Cc/Cd= 1.0 10⁻⁴ M/Cd= 1.0 10⁻⁵ M= 10 (so, we have to dilute the solution 10 times, and for this we have to take a volume 10 times lower than the final volume).

To prepare the solution, we take 2.5 ml of 1.0 10⁻⁵ M CV+, we dispense the volume in a 25 ml volumetric flask and then we add water until complete 25 ml (aproximately 22.5 ml of water).

In the second part is the same. We have: Vd= 10 ml; Cc= 1.0 10⁻⁴M; Cd= 3.0 10⁻⁵M.

Vc= Vd x Cd / Cc= (10 ml x 3.0 10⁻⁵ M)/1.0 10⁻⁴ M = 3 ml

To prepare the solution, we take 3 ml of 3.0 10⁻⁵ M CV+, we dispense the volume in a 10 ml volumetric flask and then we add water until we complete 10 ml (aproximately 7 ml of water).

6 0

3 years ago