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2 years ago

HELP PLEASE :((( Explain why helium filled balloons deflate over time faster than air filled balloons do?

Chemistry

2 answers:

Finger [1]2 years ago

6 0

Answer:

helium filled balloons tend to deflate over time faster than air filled balloons do because, the average velocity of helium molecules is higher than that of molecules of any gas found in air, so helium escapes through the pores in the balloon more quickly.

Explanation:

after researching I found this and tried to respond as quickly as possible!! hope this helps but it has something to do with the molecules .

Shkiper50 [21]2 years ago

3 0

Answer:

Because helium has more molecules trying to escape at faster rates

Explanation:

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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

Which equation correctly defines Hooke's law?

den301095 [7]

I think it is the third one.

4 0

3 years ago

Read 2 more answers

Which of the following may indicate that a chemical reaction has occurred?

lianna [129]

Answer:

D, all of the above

Explanation:

3 0

2 years ago

(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction

ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

$Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2$

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

$m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2$

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

$m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu} \\\\m_{Cu}=0.380gCu$

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

$m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3} \\\\m_{AgNO_3}=2.03gAgNO_3$

Best regards!

5 0

3 years ago

Calculate the solubility of Mg(OH)2 in water at 25 C. You'll find Ksp data in the ALEKS Data tab. Round your answer to significa

elena-s [515]

Answer:

1.12 × 10⁻⁴ M

Explanation:

Step 1: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 2: Make an ICE chart

We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

I 0 0

C +S +2S

E S 2S

The solubility product constant is:

Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³

S = 1.12 × 10⁻⁴ M

8 0

2 years ago