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Katyanochek1 [597]

3 years ago

11

What is the atomic mass of copper?

1 answer:

kipiarov [429]3 years ago

4 0

Answer:

63.546 u

Explanation:

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Do you expect to find an atom with 26 protons and mass number of 52 in nature

PilotLPTM [1.2K]

Yes, i expect to find an atom with 26 protons and mass number of 52, because if it has different number of neutrons it could be a type of isotope. As if there is different number of neutrons in an atom, they make an isotopes of that atom and if we add the number of neutrons and number of protons we will find the mass number of that atom.

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3 years ago

Avogadro counted the number of __ in carbon 12 gas

Burka [1]

Answer:

(atoms.............) is the homogeneous mixture

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3 years ago

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Suppose that H2SO4 was used in the reaction instead of HCl. How many moles of NaOH would neutralize 1 mole of H2SO4?

arlik [135]

Answer: Hence, 2 moles of NaOH would neutralize 1 mole of $H_2SO_4$

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

One mole of hydrochloric acid neutralizes one mole of sodium hydroxide to give one mole of sodium chloride and one mole of water

When sulfuric acid is used in the place of HCl , 1 mole of sulfuric acid will neutralizes the 2 mole of sodium hydroxide and gives one mole of sodium sulfate and 2 moles of water. As we can see from the reaction:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

Hence, 2 moles of NaOH would neutralize 1 mole of $H_2SO_4$

5 0

3 years ago

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Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago

Help. 10 points per question answered and brainiest answer

Soloha48 [4]

The first one is it increases
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3 years ago