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3 years ago

How many molecules of water are present in a 5.65 mole sample of water

Chemistry

1 answer:

BARSIC [14]3 years ago

3 0

Answer:

5.65

Explanation:

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A weight of 100 pounds moved a distance of 12 feet equals 120 240 1200 ft.-lbs. of work.

Archy [21]

Work = Fdcos(alpha)

Where F is force, D is distance, and Alpha is the angle to the parallel plane (in this case the ground)

I’ll assume alpha is 0 since it’s not given

100 • 12 • cos(0) = 1200 ft-lbs

3 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

Write the balanced NET IONIC equation for the reaction that occurs when ammonium nitrate and potassium hydroxide are combined. N

vova2212 [387]

Answer:

Net Ionic equation

NH₄⁺ + OH⁻ → NH₃ + H₂O

Option B is correct.

Weak Acid Strong Base

Check Explanation for the extent of the reaction.

Explanation:

Ammonium nitrate = NH₄NO₃

Potassium Hydroxide = KOH

Ammonium salts combine with alkalis to liberate NH₃ and form water.

The two reactants combine to give

NH₄NO₃ + KOH → KNO₃ + NH₃ + H₂O

In ionic form,

- NH₄NO₃ exists as NH₄⁺ and NO₃⁻

- KOH exists as K⁺ and OH⁻

- KNO₃ as K⁺ and NO₃⁻

And NH₃ and H₂O stay as they are, as per covalent compounds.

So, we have

NH₄⁺ + NO₃⁻ + K⁺ + OH⁻ → K⁺ + NO₃⁻ + NH₃ + H₂O

Eliminating the ions that exist on both sides, we have the net ionic equation to be

NH₄⁺ + OH⁻ → NH₃ + H₂O

which shows that this reaction is essentially a neutralization reaction in which the Bronsted Lowry acid, NH₄⁺, loses its proton to the base, OH⁻ and gives conjugate base, NH₃ and conjugate acid, H₂O.

This reaction is classified as a Weak acid versus Strong Base reaction as NH₄⁺ is from a Weak acid and OH⁻ is from a strong base.

Since this reaction is between a Weak base and a strong acid, the ionization isn't expected to be 100%, Hence, the extent of this reaction will be any option that is not 100%, a couple pieces of information might be required for the correct estimate, but above 50% seems correct.

Hope this Helps!!!

8 0

3 years ago

Calculate the mass of hydrogen in 500 g of Al(C2H3O2)3

Zigmanuir [339]

Answer is: the mass of hydrogen is 22,05 grams.
m(Al(C₂H₃O₂)₃) = 500 g.
M(Al(C₂H₃O₂)₃) = 27 + 6 ·12 + 9 · 1 + 6 · 16 · g/mol = 204 g/mol.
n(Al(C₂H₃O₂)₃) = m(Al(C₂H₃O₂)₃) ÷ M(Al(C₂H₃O₂)₃).
n(Al(C₂H₃O₂)₃) = 500 g ÷ 204 g/mol.
n(Al(C₂H₃O₂)₃) = 2,45 mol.
n(Al(C₂H₃O₂)₃) : n(H) = 1 : 9.
n(H) = 22,05 mol.
m(H) = 22,05 mol · 1 g/mol
m(H) = 22,05 g.

3 0

3 years ago

Read 2 more answers

Please help with chemistry homework (I’ll mark brainliest!)

masha68 [24]

Answer:

You answer is correct

Electrons have a negative charge. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.

Thank you and please rate me as brainliest as it will help me to level up

6 0

3 years ago