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Juli2301 [7.4K]

4 years ago

8

Charcoal is primarily carbon. what mass of co2 is produced if you burn enough carbon (in the form of charcoal) to produce 4.80 k

j×102kj of heat? the balanced chemical equation is as follows:

1 answer:

Reika [66]4 years ago

6 0

When carbon is burned in air carbon iv oxide gas is formed.
C (s) + O2 (g) = CO2(g) ΔH = - 393.5 kj/mol
The enthalpy change of the reaction is -393.5 j/mol which means that when one mole of carbon is completely burnt in air then 393.5 j of energy is evolved.
Thus, 1 mole = -393.5 j , then for 480 kj
= 480 × 1/393.5
= 1.2198 moles
1 mole of carbon iv oxide is equal to 44 g
thus, 1.2198 moles will be 1.2198 × 44 = 53.6712 g of CO2

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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

What volume of 0.20 M solution of HCI will be required to neutralize 10.0 mL of a 0.60 M

tatyana61 [14]

Answer:

30ml

Explanation:

Chcl.Vhvl=Ckoh.Vkoh

4 0

3 years ago

The combination of oxygen with other substances to produce new chemical products is called

Flauer [41]

Answer:

The combination of oxygen with other substances to produce new chemical products is called <u>Oxidation</u>.

Explanation:

Oxidation reactions are defined as,

In terms of Inorganic chemistry:

(i) <u>Removal of Electrons: </u>

Example: Mg → Mg²⁺ + 2 e⁻

(ii) <u>Addition of Oxygen:</u>

Example: 2 Mg + O₂ → 2 MgO

In terms of Organic chemistry:

(i) <u>Addition of Electrons: </u>

Example: Cl₂ + 2 e⁻ → 2 Cl⁻

(ii) <u>Addition of Hydrogen:</u>

Example: H₂CCH₂ + H₂ → H₃CCH₃

4 0

3 years ago

A rigid container of O has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273 K?

tia_tia [17]

Answer:

P₂ = 130.18 kPa

Explanation:

In this case, we need to apply the Gay-Lussack's law assuming that the volume of the container remains constant. If that's the case, then:

P₁/T₁ = P₂/T₂ (1)

From here, we can solve for the Pressure at 273 K:

P₂ = P₁ * T₂ / T₁ (2)

Now, all we need to do is replace the given data and solve for P₂:

P₂ = 340 * 273 / 713

<h2>P₂ = 130.18 kPa</h2>

Hope this helps

4 0

3 years ago

Suppose you are provided with a 30.86 g sample of potassium chlorate to perform this experiment. What is the mass of oxygen you

oee [108]

Answer:

The mass of oxygen is 12.10 g.

Explanation:

The decomposition reaction of potassium chlorate is the following:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

We need to find the number of moles of KClO₃:

$\eta_{KClO_{3}} = \frac{m}{M}$

Where:

m: is the mass = 30.86 g

M: is the molar mass = 122.55 g/mol

$\eta_{KClO_{3}} = \frac{30.86 g}{122.55 g/mol} = 0.252 moles$

Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3

$\eta_{O_{2}} = \frac{3}{2}*0.252 moles = 0.378 moles$

Finally, the mass of O₂ is:

$m = 0.378 moles*32 g/mol = 12.10 g$

Therefore, the mass of oxygen is 12.10 g.

I hope it helps you!

6 0

3 years ago