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3 years ago

Why are metal containers not used for storing acids

1 answer:

3 0

Answer : Metal containers are not used for storing acid because most of the time acid reacts with almost every metal and produces salts or oxides which alters the acid characteristics making it useless .......

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An 85 kg construction worker has 37 485 J of gravitational potential energy. To the nearest whole number, the worker is___

Dmitry [639]

Answer:

44,1 m

Explanation:

Ep= m * g * h

37485 = 85 * 10 * h

h = 44,1 m

7 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]

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

Explanation:

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:

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

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

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

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 $1.87\times 10^{-14}M$

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3 years ago

what would the percentage of thorium-232 ( parent compound) be in a rock that was dated at 1.8 billion years

vovikov84 [41]

The half-life of Th-232 is 1.405 × 10¹⁰ years

Time elapsed = 2.8 x 10⁹ years

Equation of radioactive decay:

A = A₀ = (1/2)^ t/t₁/₂

Thus, the percentage of thorium-232 in the rock that was dated at 2.8 billions year = 87.1%

7 0

3 years ago

If an atom of chlorine had an atomic mass of 17 and it gained an electron what would you have?

Sonbull [250]

The atomic mass would not change since the mass of an electron is negligible compared to the mass of protons and neutrons

4 0

3 years ago

A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 da

Flura [38]

55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g

7 0

3 years ago

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Why are metal containers not used for storing acids​

Why are metal containers not used for storing acids