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2 years ago

What are household items that are made up from carbon?

2 answers:

6 0

You're made partly of carbon, so is clothing, furniture, plastics and your household machines. There is carbon in the air we breathe. Diamonds and graphite are also made of carbon.Apr

sweet-ann [11.9K]2 years ago

4 0

you’re made partly of carbon so is clothes, furniture, plastics, yr household machines

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Are controlled substances and illegal drugs the same thing?

Reil [10]

Well i do think they're the same.

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3 years ago

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The questions in bold are questions you shouldnt answer, answer the rest.

velikii [3]

A variable is not consistent or having a fixed pattern; liable to change.

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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

Convert 7.34 miles into meters (1mile = 1.6 Km)

Lera25 [3.4K]

Answer:

11812.58 meters = 11.81258 Km

Explanation:

hope that helps

8 0

3 years ago

A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated cylin

gtnhenbr [62]

Answer:

ksp = 0,176

Explanation:

The borax (Na₂borate) in water is in equilibrium, thus:

Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)

When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ (1)

The ksp is defined as:

ksp = [borate²⁻] [Na⁺]²

Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:

B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻

The moles of HCl that reacts with B₄O₇²⁻ are:

0,500M×0,01200L = 6,00x10⁻³ mol of HCl

As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:

6,00x10⁻³ mol of HCl× $\frac{1molB_{4}O_{7}^{2-}}{2molHCl}$ = 3,00x10⁻³ mol of B₄O₇²⁻

For (1), moles of Na⁺ are 3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺

The [borate²⁻] is 3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = 0,353M

And [Na⁺] is 6,00x10⁻³ mol of Na⁺ / 0,00850L = 0,706M

Replacing in the expression of ksp:

ksp = [0,353] [0,706]²

ksp = 0,176

I hope it helps!

8 0

2 years ago