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2 years ago

What is the IUPAC name for this product?

Chemistry

1 answer:

mihalych1998 [28]2 years ago

7 0

The IUPAC name for the given product is 2 chloro Butane.

<h3>What is IUPAC nomenclature?</h3>

IUPAC stands for 'International Union of Pure and Applied Chemistry', which givers some rule for designing the name of compounds of chemistry.

In the given product total four carbon atoms are present and between all of them single bonds are present.
In the second carbon atom, chlorine group is present.
During the nomenclature process, first we write down the name of the attached group which is followed by the alkane chain.

Hence name of the product is 2 chloro Butane.

To know more about IUPAC nomenclature, visit the below link:

brainly.com/question/26635784

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How to clean a nose piercing with antibacterial soap?.

pychu [463]

Cleaning the nose piercing with a few simple steps will help prevent pre-existing conditions like infections, allergic reactions or skin damage. Performing regular maintenance will keep your piercing cleaner, more comfortable and easier to clean in the future.(Remember: If you have any problems I recommend visiting a doctor at your earliest convenience!) Antibacterial soap is great at reducing inflammation and clearing up infections. You should soak your nose in it regularly and use a clean cloth to gently wash away any particles that may have fallen into the septum.

5 0

2 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

3 years ago

What is the complete ionic equation for NaOHaq plus HCIaq equals H2OI plus NaCIaq?

OverLord2011 [107]

Answer:

Ionic equation:

Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻ (aq)

Explanation:

Chemical equation:

NaOH(aq) + HCl(aq) → H₂O(l) + NaCl (aq)

Balanced chemical equation:

NaOH(aq) + HCl(aq) → H₂O(l) + NaCl (aq)

Ionic equation:

Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻ (aq)

Net ionic equation:

OH⁻(aq) + H⁺(aq) → H₂O(l)

The Cl⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

3 0

3 years ago

Help me out w this plz

zhannawk [14.2K]

Cant see the picture

4 0

3 years ago

A non-stoichiometric compound is a compound that cannot be represented by a small whole-number ratio of atoms, usually because o

ikadub [295]

<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38

Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238
The total charge of all the vanadium ions would be +238.
Let x = number of of V+2
Then 100 – x = number of V+3
X(+2) + 100-x(+3) = +238
2x + 300 – 3x = 238
-x = 238-300 = -62
x = 62

Thus 62/100 are V+2
62/100 * 100 = 62%

</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

7 0

3 years ago