Why is dissolving salt in water is a physical change? Help me please

How do you know when a phase change happens?

VladimirAG [237]

Answer:

phase for what?

Explanation:

4 0

3 years ago

Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that

Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻ E° = 2.38 V oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s) E° = -0.25 V reducting

------------------------------------------------------------

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s) E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0

3 years ago

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-

hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

First Formula would be $ZnF_{2}$

Second empirical formula is between $Zn^{2+}(Positive ion) and O^{2-}(Negative ion)$

Second Formula would be $Zn_{2}O_{2}$

Note : When the subscript are same they get cancel out, so $Zn_{2}O_{2}$ would be written as $ZnO$

Third empirical formula is between $Ni^{4+}(Positive ion) and F^{-}(Negative ion)$

Third Formula would be : $NiF_{4}$

Forth empirical formula is between $Ni^{4+}(Positive ion)and O^{2-}(negative ion)$

Forth Formula would be : $Ni_{2}O_{4}$ or $NiO_{2}$

Note- The subscript will be simplified and the formula will be written as $NiO_{2}$ .

The empirical formula of four binary ionic compounds are : $ZnF_{2}, ZnO, NiF_{4},NiO_{2}$

8 0

3 years ago

Read 2 more answers

Hello! I just need a little bit of help. I'm supposed to design an experiment on how reaction rates are determined and affected

alexandr1967 [171]

Get 3 cups of water at the exact same temperature, using the thermometer to check.
Label the cups as ‘whole’, ‘pieces’, and ‘crushed’
Next, get something to dissolve, in this case, polident. Take one of the polident tablets and break it into 4 pieces, and set it aside.
Take another polident tablet and this time put it into a different cup, and crush it. Set it aside.
Keep the last tablet whole.
Set up your stopwatch and drop the polident tablet that is whole in the cup labeled ‘whole’, starting the stopwatch at the same time.
Watch the cup and see when the tablet is fully dissolved, then stop the stopwatch.
Record the time in the table.
Repeat steps 6-8 for both the ‘pieces’ and ‘crushed’ tablets.

Hope this helps! Please let me know if you need more help, or if you think my answer is incorrect. Brainliest would be MUCH appreciated. Have a great day!

Stay Brainy!

− $xXheyoXx$

3 0

2 years ago

The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at

nalin [4]

Answer:

The given statement - The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at each other along the line between the two nuclei , is <u>True.</u>

Explanation:

The above statement is correct , because the sigma bond is produced by the head on overlapping, the orbitals should all point in the same direction.

<u>SIGMA BONDS -</u> Sigma bonds (bonds) are the strongest type of covalent chemical bond in chemistry. They're made up of atomic orbitals that collide head-on. For diatomic molecules, sigma bonding is best characterized using the language and tools of symmetry groups.

Head-on overlapping of atomic orbitals produces sigma bonds. The concept of sigma bonding is expanded to include bonding interactions where a single lobe of one orbital overlaps with a single lobe of another. Propane, for example, is made up of ten sigma bonds, one for each of the two CC bonds and one for each of the eight CH bonds.

Hence , the answer is true .

6 0

3 years ago