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FrozenT [24]
2 years ago
11

4. Suppose you want to deposit silver on an iron spoon using silver nitrate as

Chemistry
1 answer:
Levart [38]2 years ago
8 0

The spoon should be connected to the negative terminal of the battery. The other electrode should be made of silver.

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3) A sample of chlorine gas has a volume of 23.234 Lat STP. How many moles is this?
andreyandreev [35.5K]

Answer:

Explanation:

jordan I am literally stuck on the same problem LOLLL WE REALLY OUT HERE

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3 years ago
Which of these phrases would go in the overlap? Select two options.
SVETLANKA909090 [29]
I think it’s found in the nucleus and has a mass of one amu
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1 year ago
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Given the following reaction: 3D(g) + E(g) + 2F(g) → 5G(g) + 4H(g)
evablogger [386]

Answer:

Explanation:

Hello,

The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:

\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}

Now, we answer:

(a)

\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s

(b)

\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s

(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:

r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}

Thus, any of the available expressions are suitable to quantify the rate of the reaction.

Best regards.

4 0
3 years ago
Which of the following is an electrolyte?
Serggg [28]
Electrolytes are inorganic substances that dissociate into ions in water. Examples of electrolytes include salts [for example, sodium chloride (NaCl) and magnesium chloride (MgCl2)] and ions [for example, potassium (K+)].
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3 years ago
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Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5). express your answer numerically to four s
Slav-nsk [51]
Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) = 5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.
6 0
3 years ago
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