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2 years ago

Study the figures below which illustrate the steps in the following chemical reaction:

Chemistry

1 answer:

Tanya [424]2 years ago

7 0

Answer:

Hello - this is Mrs. Gussman, your chemistry teacher. I wrote this exam question and posting it online is a violation of the academic integrity policy. Remove this post immediately.

Explanation:

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What type of glassware is useful for transferring liquid chemicals?

julia-pushkina [17]

In order to transfer liquid we need

a) a funnel: an apparatus which transfers liquid from one glassware to other (of narrow mouth) without spilling

b) We should use a beaker (a glassware with a beak for proper transfer) to transfer liquid from it to other glass apparatus.

c) other things are like transparent clear dry and clean apparatus should be used

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4 years ago

Many drugs are sold as their hydrochloric salts (R2NH2+Cl−), formed by reaction of an amine (R2NH) with HCl. Part 1 out of 4 Dra

Drupady [299]

Answer:

Acebutolol hydrochloride is the form of the hydrochloride salt of acebutolol, a synthetic derivative of butyranide with a hypotensive and antiarrhythmic activity. Acebutolol acts as a cardioselective beta-adrenergic antagonist with very little effect on bronchial receptors, having intrinsic sympathomimetic properties. Acebutolol is used in ventricular arrhythmias. Other indications may include hypertension, alone or in combination with other drugs. The salt scheme is found in the attached file.

Explanation:

Download docx

4 0

3 years ago

What role does the salt bridge play in a voltaic cell? It allows electrons to flow from the anode to the cathode.It allows migra

Mnenie [13.5K]

The answer is it contains the electrodes. Without the salt scaffold, the arrangement in the anode compartment would turn out to be decidedly charged and the arrangement in the cathode compartment would turn out to be contrarily charged, on account of the charge lopsidedness, the terminal response would rapidly stop.
It keeps up the stream of electrons from the oxidation half-cell to a decrease half cell, this finishes the circuit.

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4 years ago

The pH of a solution is 8.83±0.048.83±0.04 . What is the concentration of H+H+ in the solution and its absolute uncertainty?

slava [35]

Answer:

The concentration of $H^{+}$ is 1.48 × $10^{-9}$ M

The absolute uncertainty of $[{H^{+}]$ is ±0.12 × $10^{-9}$ M

The concentration of $H^{+}$ is written as 1.48(±0.12) × $10^{-9}$ M

Explanation:

The pH of a solution is given by the formula below

pH = $-log_{10}[{H^{+}]$

∴ $[H^{+}] = 10^{-pH}$

where $[{H^{+}]$ is the $H^{+}$ concentration

From the question,

pH = 8.83±0.04

That is,

pH =8.83 and the uncertainty is ±0.04

First, we will determine $[{H^{+}]$ from

$[H^{+}] = 10^{-pH}$

$[{H^{+}] = 10^{-8.83}$

$[{H^{+}] = 1.4791$ × $10^{-9}$ M

$[{H^{+}] = 1.48$ × $10^{-9}$ M

The concentration of $H^{+}$ is 1.48 × $10^{-9}$ M

The uncertainty of $[{H^{+}]$ ( $U_{[H^{+}] }$ ) from the equation $[H^{+}] = 10^{-pH}$ is

$U_{[H^{+}] } = 2.303 \\$ × ${[H^{+}] }$ × $U_{pH }$

Where $U_{[H^{+}] }$ is the uncertainty of $[{H^{+}]$

$U_{pH }$ is the uncertainty of the pH

Hence,

$U_{[H^{+}] }$ = 2.303 × 1.4791 × $10^{-9}$ × 0.04

$U_{[H^{+}] }$ = 1.36 × $10^{-10}$ M

$U_{[H^{+}] }$ = 0.12 × $10^{-9}$ M

Hence, the absolute uncertainty of $[{H^{+}]$ is ±0.12 × $10^{-9}$ M

6 0

4 years ago

It is expected that a chemical reaction will occur when copper metal is combined with aqueous zinc sulfate. Explain why there wi

REY [17]

Answer:

$E_{cell}$ = +ve, reaction is spontaneous

$E_{cell}$ = -ve, reaction is non spontaneous

$E_{cell}$ = 0, reaction is in equilibrium

Case 1: when copper metal is combined with aqueous zinc sulfate.

$Cu+ZnSO_4\rightarrow Zn+CuSO_4$

Here copper is undergoing oxidation ad thus acts as anode and zinc is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Cu^{2+}/Cu]}$

$E^0=-0.76-(+0.34)=-1.10V$

Thus as $E_{cell}$ is negative , the reaction is non spontaneous.

Case 2: when zinc metal and aqueous copper sulfate solution are combined.

$Zn+CuSO_4\rightarrow Cu+ZnSO_4$

Here zinc is undergoing oxidation ad thus acts as anode and copper is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential = $E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Zn^{2+}/Zn]}$

$E^0=+0.34-(-0.76)=+1.10V$

Thus as $E_{cell}$ is positive , the reaction is spontaneous.

3 0

4 years ago