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2 years ago

Help guys please question 3 in picture ASAP pls

Chemistry

1 answer:

vazorg [7]2 years ago

8 0

Answer:

1 Spinal fluid

2 Milk

3 Saliva

4 Urine

5 Gastric content

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to calculate the pH of both gastric content and spinal fluid by using the following equations and works:

$pH_{gastric}=-log(10^{-2})=2.0\\\\pH_{spinal}=14+log(10^{-6.6})=7.4$

Thus, we rank them as follows:

1 Spinal fluid

2 Milk

3 Saliva

4 Urine

5 Gastric content

Regards!

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A certain first-order reaction (A→products) has a rate constant of 7.20×10−3 s−1 at 45 ∘C. How many minutes does it take for the

lesya692 [45]

Answer:

6,41 min

Explanation:

For the reaction:

A → products

kinetics first-order reaction law is:

ln[A] = ln[A]₀ -kt

Where [A] is concentration of reactant, [A]₀ is initital concentration of reactant, k is rate constant and t is time.

If the concentration of A is 6,25% you can assume:

[A] = 6,25; [A]₀= 100. Replacing:

ln(6,25) = ln(100) -7,20×10⁻³s⁻¹t

-2,7726 = -7,20×10⁻³s⁻¹t

385s = t

In minutes:

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I hope it helps!

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3 years ago

A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a SoaveRedlich-Kwong gas and with

Setler79 [48]

Explanation:

The given data is as follows.

Moles of propylene = 100 moles, $T_{i}$ = 300 K

$T_{f}$ = 800 K, $V_{i} = 2 m^{3}$

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Now, we assume the following assumptions:

Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.

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3 years ago

Why is a wire used to connect the electrodes in an electrochemical cell?

jarptica [38.1K]

Answer:

D. So electrons can flow from one electrode to another

Explanation:

An electrochemical cell is any cell in which electricity is produced by reason of a chemical change.

An electrochemical cell consists of two electrodes, these two electrodes are connected using a wire.

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3 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

(a) As per question, lead is oxidized and copper is reduced.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

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$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

(b) As per question, lead is reduced and copper is oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

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

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$