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3 years ago

What is the pH of the lye in soap ?

Chemistry

2 answers:

lord [1]3 years ago

7 0

Answer:

The pH of the lye in soap is 13.

Explanation:

igomit [66]3 years ago

3 0

9-10, when too much fluke is used there is not enough oil to turn it into soap. This lye is “free-floating” in the soap which increases the pH level. Soap with high pH will irritate or burn the skin and should not be used.

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The question is in the picture below.

pantera1 [17]

The correct option is A.
An oxidation reaction is one in which a substance gives away electrons and becomes oxidized. In the equation given above, the chlorate ion undergoes oxidation reaction and gives away two chlorine ion.

4 0

3 years ago

Why doesn't potassium fluoride conduct electricity as a solid?

Korvikt [17]

potassium flouride is an ionic compound and ionic compounds do not conduct electricity in solid form instead it conducts electricity in molten state.Ionic compounds do not conduct electricity in solid state because their ions are fixed while in molten form the ions are free

5 0

3 years ago

For the following reactions, predict the products and write the balanced formula equation, complete ionic equation, and net ioni

stealth61 [152]

Answer:

Explanation:

To predict the products of these reactions we need to know the kind of reactions. All these reactions are double replacement reaction. In these kinds of reactions, the products will be the result of exchanging ions in the reactants. So, the first step is to identify the ions.

For the reaction, we have Hg2(NO3)2 and CuSO4. We have the ions Hg+1, NO3-1, Cu+2 and SO4-2

The way to make this exchange is putting together positive in one species with the negative of the other species. Following that rule we have

$Hg^{+1} - - - (SO_{4})^{-2}[/text]
the oxidation number will tell you the subscript for each species in the compound. In this case, is Hg2(SO4) [tex]Cu^{+2} - - - (NO_{3})^{-1} - - -> Cu(NO_{3})_{2} [/text] 
So, the products for this reaction will be
 [tex]Hg_{2} (NO_{3})_{2}(aq) + CuSO_{4}(aq) --> Hg_{2}SO_{4} + Cu(NO_{3})_{2}[/text]

After this, we proceed to balance the equation. For this, we check that we have the same number of each element on both sides of the equation. In this case, we can see that we have the same number, so the equation is balanced. Finally, we check the rules of solubility to see if the species are soluble in water or not. In this case sulfates area always soluble except for mercury so Hg2(SO4) precipitates in the solution (pre). Nitrates are always soluble so Cu(NO3)2 is soluble (aq) 
[tex] Hg_{2}(NO_{3})_{2}(aq) + CuSO_{4}(aq) - -> Hg_{2}SO_{4} (pre) + Cu(NO_{3})_{2}(aq)$

The complete ionic equation allows to show which of the reactants or products exist primarily as ions. For this reaction this will be:

$2Hg^{+1}(aq) + 2(NO_{3})^{-1}(aq) + (SO_{4})^{-2}(aq) + Cu^{+2}(aq) --> Hg_{2}SO_{4} (pre)+ Cu^{+2}(aq) + (NO_{3})^{-1}(aq) [/text]

To get net ionic equation we take away the ions that did not participate in the reactions. In other words the ones that are the same on both sides in the equation. In this case we see that [tex] Cu^{+2}(aq) and (NO_{3})^{-1}(aq) [/text] are the same on both sides so those ions are not include in the net ionic equation. This is:
[tex] 2Hg^{+1}(aq) + (SO_{4})^{-2}(aq) --> Hg_{2}SO_{4} (pre) [/text]

B [tex] Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq)$

ions (1) $Ni^{+2} and (NO_{3})^{-1}$

ions (2) $Ca^{+2} and Cl^{-1}$

Exchanging

$Ni^{+2} ---- Cl^{-1} --> NiCl_{2}$

$Ca^{+2} --- (NO_{3})^{-1} --> Ca(NO_{3})_{2}$

Products

$Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq) --> NiCl_{2} + Ca(NO_{3})_{2}$

The equation is already balanced

Chlorides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NiCl2 is soluble (aq)

Nitrates are always soluble. Ca(NO3)2 is soluble (aq)

Since both compounds are soluble, we can say that there is not reaction.

Complete ionic equation

$Ni^{+2}(aq) + 2(NO_{3})^{-1} (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq) - - > Ni^{+2}(aq) + 2(NO_{3})^{-1} (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq)$

Net ionic equation:

The ions in both sides of the equation are the same so all of them are cancelled and we cannot get a net ionic equation this explains why there is no reaction in this case.

C $K_{2}CO_{3}(aq) + MgI_{2}(aq)$

Ions(1) $K^{+1} and (CO_{3})^{-2}$

Ions(2) $Mg^{+2} and l^{-1}$

Exchanging

$K^{+1} --- l^{-1} - - > KI$

$Mg^{+2} --- (CO_{3})^{-2} - - > Ca(CO_{3})$

Products

$K_{2}CO_{3}(aq) + MgI_{2}(aq) - -> Kl + MgCO_{3}$

The equation is not balanced

Balance equation is

$K_{2}CO_{3}(aq) + MgI_{2}(aq) - -> 2Kl (aq) + MgCO_{3} (pre)$

iodides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. KI is soluble (aq)

carbonates are always insoluble except group 1 cations. MgCO3 is insoluble (pre)

complete ionic equation

$2K^{+1}(aq) + (CO_{3})^{-2}(aq) + Mg^{+2}(aq) + 2l^{-1}(aq) - - > MgCO_{3} (pre) + 2K^{+1}(aq) + 2l^{-1}(aq)$

Net ionic equation

$(CO_{3})^{-2}(aq) + Mg^{+2}(aq) - - > MgCO_{3} (pre)$

D $Na_{2}CrO_{4}(aq) + AlBr_{3}(aq)$

Ions(1) $Na^{+1} and (CrO_{4})^{-2}$

Ions(2) $Al^{+3} and Br^{-1}$

Exchanging

$Na^{+1} ---- Br^{-1} - -> NaBr$

$Al^{+3} --- (CrO_{4})^{-2} - -> Al_{2}(CrO_{4})_{3}$

Products

$Na_{2}CrO_{4}(aq) + AlBr_{3}(aq) - -> NaBr + Al_{2}(CrO_{4})_{3}$

The equation is not balanced

Balance equation is

$3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr + Al_{2}(CrO_{4})_{3}$

bromides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NaBr is soluble (aq)

chromates are always insoluble except group 1 cations. Al2(CrO4)3 is insoluble (pre)

$3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr(aq) + Al_{2}(CrO_{4})_{3}(pre)$

Complete ionic equation

$6Na^{+1}(aq) + 3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) + 6Br^{-1}(aq) - -> Al_{2}(CrO_{4})_{3}(pre) +6Br^{-1}(aq) + 6Na^{+1}(aq)$

Net ionic equation

$3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) - -> Al_{2}(CrO_{4})_{3}(pre)$

6 0

3 years ago

Which example best describes a reflex action?

Pepsi [2]

I think the answer would be b because your throat is reacting to the cough?

6 0

3 years ago

Use the periodic table to determine the electron configuration for Ca and Pm in noble-gas notation Ca: [Ar]4s2 [Ar]4s1 [Ar]3s2 [

sveta [45]

Answer:

1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵

Explanation:

1) Ca:

Its atomic number is 20. So it has 20 protons and 20 electrons.

Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.

So, you have two more electrons (20 - 18 = 2) to distribute.

Those two electrons go the the orbital 4s.

Finally, the electron configuration is [Ar] 4s².

2) Pm

The atomic number of Pm is 61, so it has 61 protons and 61 electrons.

Pm is in the row (period) 6. So, the noble gas before Pm is Xe.

The atomic number of Xe is 54.

Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.

The resultant distribution for Pm is: [Xe]6s² 4f⁵.

5 0

3 years ago

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