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3 years ago

12

Table salt is considered a(n) ______________ because it conducts electricity in water. A) electrolyte B) electrolysis C) nonelec

trolyte D) electrochemical cell Elimi

2 answers:

NikAS [45]3 years ago

7 0

The answer to this question would be A.)electrolyte

kodGreya [7K]3 years ago

4 0

A) electrolyte the others either describe a process, the opposite, or a whole different thing

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How is scientific observation done properly and safely using laboratory equipment?

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3 years ago

What is the volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm

Virty [35]

The volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm is 11.06L.

<h3>How to calculate volume?</h3>

The volume of a given mass of gas can be calculated using the following formula:

PV = nRT

Where;

P = pressure
V = volume
R = gas law constant
T = temperature
n = number of moles

According to this question, 0.98 moles of oxygen gas at 275 k contains a pressure of 2.0 atm. The volume is calculated as follows:

2 × V = 0.98 × 0.0821 × 275

2V = 22.13

V = 11.06L

Therefore, the volume of 0.98 mol oxygen gas at 275 k and a pressure of 2.0 atm is 11.06L.

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2 years ago

roundup an herbicide manufactured by monsanto has the formula c3h8no5p. how many moles of molecules are there in a 500g sample o

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3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago