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KonstantinChe [14]
3 years ago
10

Is granite rock a compound, element, mixture? Explain

Chemistry
1 answer:
Ghella [55]3 years ago
7 0
It is a <u>mixture</u> of minerals mainly of corps that crystalizes of magma below earths surface.
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Which of the following is an ionic compound whose aqueous solution conducts electricity?
Stels [109]
B.) ELECTROLYTE..............
4 0
3 years ago
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
Which is an example of gene modification that humans have done for thousands of years?
miv72 [106K]
Cross breeding plants to make better apples
4 0
3 years ago
Can somebody help? Do not answer if you don't know.
blondinia [14]

Answer:

10.8amu

Explanation:

Given parameters:

Abundance of B - 10  = 20% = 0.2

Abundance of B - 11  = 80% = 0.8

Unknown:

Atomic mass of Boron = ?

Solution:

The atomic mass of Boron can be can be calculated using the expression below;

 Atomic mass  = (abundance of B - 10 x mass of isotope B - 10 ) +( abundance of  B - 11 x mass of isotope B- 11)

 Atomic mass  = (0.2 x 10) + (0.8 x 11) = 2 + 8.8  = 10.8amu

5 0
2 years ago
How many moles of N are in 0.221 g of N2O?
SashulF [63]
No.of moles of N2O= 0.221g/(14*2+16) =0.005022727moles
no.of moles of N = 0.005022727*2 =0.010045454moles
4 0
3 years ago
Read 2 more answers
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