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4 years ago

What is an example of a colloid? A. Soda B.Fog C.Jewelry D.Silverware

Chemistry

2 answers:

TEA [102]4 years ago

5 0

Answer:

B. fog

Explanation:

This is because a liquid (water) is dispersed throughout the air (gas)

Olegator [25]4 years ago

4 0

Answer:

the current answer is

silverware

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How many electrons does a single oxygen gain or lose in the following reaction?

REY [17]

Answer:

it gains 2 electrons

Explanation:

atoms want to be more stable, for oxygen it's number is 8, 2 less than the stable ring of 10. and hydrogen has a single electron, 1 less than the stable ring of 2. so oxygen takes 2 electrons 2 make it stable and hydrogen becomes attatched to the oxygen atom to form a covalent bond

8 0

3 years ago

Read 2 more answers

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

4 years ago

Calculate the volume in liters of a 29.8 g/dL nickel(II) chloride solution that contains 131. G of nickel(II) chloride . Be sure

ohaa [14]

Answer:

0.44 L.

Explanation:

Density of nickel(II) chloride = 29.8 g/dL.

Mass of nickel(II) chloride = 131 g

Volume of nickel(II) chloride =?

Next, we shall convert 29.8 g/dL to g/L. This can be obtained as follow:

Recall:

1 g/dL = 10 g/L

Therefore,

29.8 g/dL = 29.8 x 10 = 298 g/L

Therefore, 29.8 g/dL is equivalent to 298 g/L.

Finally, we shall determine the volume of nickel(II) chloride as follow:

Density of nickel(II) chloride = 298 g/L

Mass of nickel(II) chloride = 131 g

Volume of nickel(II) chloride =?

Density = mass /volume

298 = 131/Volume

Cross multiply

298 x Volume = 131

Divide both side by 298

Volume = 131/298

Volume = 0.44 L

Therefore, the volume of of nickel(II) chloride is 0.44 L

3 0

3 years ago

Penicillinase, also known as β‑lactamase, is a bacterial enzyme that hydrolyzes and inactivates the antibiotic penicillin. Penic

lana66690 [7]

Answer : The initial reaction velocity $V_o$ would approach $V_{max}$

Explanation :

According to Michaelis-Menten kinetics,

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

where,

S = substrate concentration = $10.4 \times 10^{-6} M$

$V_{max} = 6.8 \times 10^{-10} \mu mol/min$

$K_{m} = 5.2 \times 10^{-6} M$

$V_o$ = initial reaction velocity = ?

Now put all the given values into the above formula, we get:

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]$

$V_{o} = 4.5 \times 10^{-10} \mu mol/min$

Therefore, the initial reaction velocity $V_o$ would approach $V_{max}$

3 0

3 years ago

Laboratory work often involves making dilutions of standard solutions. Dilution problems are not hard, but they can sure get you

Paladinen [302]

Answer:

0.06M

Explanation:

Using the formula as outlined in this question,

C1V1= C2V2

Where;

V1 = initial volume (Litres)

C1 = initial concentration (M)

V2 = final volume (Litres)

C2 = final concentration (M)

According to the information provided on NaOH in this question, V1 = 100mL = 100/1000 = 0.1 L, V2 = 1L, C1 = 0.6M, C2 = ?

C1V1= C2V2

0.6 × 0.1 = C2 × 1

0.06 = C2

C2 = 0.06M

The new concentration of NaOH is 0.06M.

6 0

3 years ago