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laiz [17]
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
Chemistry
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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Deer! They are herbivores:))
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How is scientific observation done properly and safely using laboratory equipment?
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Using the proper protective equipment from goggles to lab coat and gloves. Also includes ventilation fan to prevent toxic fumes and emergency eyewash station
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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.

Learn more about volume at: brainly.com/question/12357202

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6 0
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
uranmaximum [27]
N= m/Mr Find the Mr of C3H8NO5P and then divide the mass by the Mr. Sorry I don't have periodic table with me so I can't calculate the Mr for you
3 0
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
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