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3 years ago

What is the the device used to measure the average kinetic energy of the particles in a substance

1 answer:

7 0

A thermometer is used to measure it because the scientific definition of temperature is the average kinetic energy of individual particles in a substance

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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

3 years ago

A sample containing 16.93 metal pellets is poured into a graduated cylinder containing 11.2 of water, causing the water level in

mixer [17]

Answer:

<h3>The answer is 1.99 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 16.93 g

volume = final volume of water - initial volume of water

volume = 19.7 - 11.2 = 8.5 mL

We have

$density = \frac{16.93}{8.5} \\ = 1.9917647058...$

We have the final answer as

Hope this helps you

7 0

3 years ago

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben

Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L = 1000 mL

1:200 solution implies the $\frac{weight}{volume}$ in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be $\frac{1000mL}{200mL}=\frac{1g}{xg}$

200mL × 1g = 1000 mL × x(g)

x(g) = $\frac{200mL*1g}{1000mL}$

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ $\frac{10mL}{250mL}=\frac{0.2g}{y(g)}$

y(g) = $\frac{250mL*0.2g}{10mL}$

y(g) = 5g of benzalkonium chloride.

Now, at 17% $\frac{weight}{volume}$ concentrate contains 17g/100ml:

∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= $\frac{17g}{5g} = \frac{100mL}{z(mL)}$

z(mL) = $\frac{100mL*5g}{17g}$

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

4 0

3 years ago

Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: when you answer the question,

vodomira [7]

Answer:

Explanation:

The result will be affected.

The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).

Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.

During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.

Thus, NaOH will appear at a higher concentration than it actually is.

7 0

3 years ago

Making methanol the element hydrogen is not abundant in nature, but it is a useful reagent in, for example, the potential synthe

stealth61 [152]

ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K

5 0

2 years ago