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3 years ago

I need you to solve for me plzzzzzz

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

3 0

Answer:

.0556 L

Explanation:

First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.

Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).

Finally, the answer will be .0556 L.

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Difference between emprical and molecular formula

Musya8 [376]

Answer:

The key difference between empirical and molecular formulas is that an empirical formula only gives the simplest ratio of atom whereas a molecular formula gives the exact number of each atom in a molecule.

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3 years ago

In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

$\text{NH}_4\text{Cl}$ is a salt soluble in water. $\text{NH}_4\text{Cl}$ dissociates into ions completely when dissolved.

$\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq)$ .

The first test tube used to contain $\text{NH}_4\text{OH}$ . $\text{NH}_4\text{OH}$ is a weak base that dissociates partially in water.

$\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+} \;(aq)+ {\text{OH}}^{-} \; (aq)$ .

There's also an equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$ .

$\text{OH}^{-}$ ions from $\text{NH}_4\text{OH}$ will shift the equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ to the right and reduce the amount of ${\text{H}_3\text{O}}^{+}$ in the solution.

The indicator equilibrium will shift to the right to produce more ${\text{H}_3\text{O}}^{+}$ ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding $\text{NH}_4\text{Cl}$ will add to the concentration of ${\text{NH}_4}^{+}$ ions in the solution. Some of the ${\text{NH}_4}^{+}$ ions will combine with $\text{OH}^{-}$ ions to produce $\text{NH}_4\text{OH}$ .

The equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions will shift to the left to produce more of both ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$

The indicator equilibrium will shift to the left as the concentration of ${\text{H}_3\text{O}}^{+}$ increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

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3 years ago

G what is the difference between the sidechains of leucine and isoleucine? select one:

statuscvo [17]

c. Isoleucine has a carbon “branched” closer to the alpha carbon than does leucine.

The structure of leucine is CH3CH(CH3)CH2CH(NH2)COOH.

The structure of isoleucine is CH3CH2CH(CH3)CH(NH2)COOH.

In leucine, the CH3 group is two carbons away from the α carbon; in isoleucine, the CH3 group is on the carbon next to the α carbon.

Thus, isoleucine has the closer branched carbon.

“One is charged, the other is not” is incorrect. Both compounds are uncharged.

“One has more H-bond acceptors than the other” is incorrect. Each acid has two H-bond acceptors — the N in the amino and the O in the carbonyl group.

“They have different numbers of carbon atoms” is incorrect. They each contain six carbon atoms.

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3 years ago

What type of pollution in cities composed of car exhaust and industrial pollutes

Troyanec [42]

Air pollution is the answer

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3 years ago

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The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell

tankabanditka [31]

Explanation:

In a voltaic cell, oxidation reaction occurs at anode whereas reduction reaction occurs at the cathode.

Hence, the half-cell reaction taking place at anode and cathode will be as follows.

At anode (Oxidation) : $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode (Reduction) : $Ag^{+}(aq) + 1e^{-} \rightarrow Ag(s)$

So, in order to balance the half cell reactions, we multiply reduction reaction by 3. Hence, reduction reaction equation will be as follows.

$3Ag^{+}(aq) + 3e^{-} \rightarrow 3Ag(s)$ ........ (2)

Therefore, overall reaction will be sum of equations as (1) + (2). Thus, net reaction equation is as follows.

$Cr(s) 3Ag^{+}(aq) \rightarrow Cr^{3+}(aq) + 3Ag(s)$

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3 years ago