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3 years ago

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You have created a way to disrupt bonds between carbon atoms and functional groups. what is the maximum number of covalent bonds

you could disrupt between six carbon atoms?

1 answer:

viva [34]3 years ago

8 0

Answer:

The carbon atom can form 4 covalent bonds to complete the eight electrons of its outermost laye, so if all the layes are completed, you will disrupt 24 bonds.

Explanation:

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PLEASE HELP!! Thanks! How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 111.0 ∘C? The

ZanzabumX [31]

Answer:

Approximately 39.7 kJ.

Assumptions: the specific heat capacity of water is $\rm 4.182\; J \cdot mol^{-1}$ , the melting point of water is $\rm 0\, ^{\circ} C$ , and that the boiling point of water is $\rm 100 \,^{\circ} C$ .

Explanation:

It takes five steps to convert 13.0 grams of $\rm \text{-}10.0\, ^{\circ}C$ ice to steam at $\rm 111.0\,^{\circ}C$ .

Step one: heat the 13.0 gram of ice from $\rm \text{-}10.0\, ^{\circ}C$ to $\rm 0\,^{\circ}C$ . The change in temperature would be $\rm 10.0\,^{\circ}C$ .
Step two: supply the heat of fusion to convert that 13.0 gram of ice to water.
Step three: heat the 13.0 gram of water from $\rm 0\,^{\circ}C$ to $\rm 100\,^{\circ}C$ . The change in temperature would be $\rm 100\,^{\circ}C$ .
Step four: supply the heat of vaporization to convert that 13.0 gram of water to steam.
Step five: heat the 13.0 gram of steam from $\rm 100\,^{\circ}C$ to $\rm 111.0\,^{\circ}C$ . The change in temperature would be $\rm 11.0\,^{\circ}C$ .

<h3>Energy required for step one, three, and five</h3>

The following equation gives the amount of energy $Q$ required to raise the temperature of an object by a $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

In this equation,

$c$ is the specific heat of this substance,
$m$ is the mass of the substance, and
$\Delta T$ is the change in the temperature of the object.

Assume that there's no mass loss in this whole process. The value of $m$ would stay the same at $13.0\; \rm g$ .

$\begin{aligned}& &&\text{Energy required for raising temperature} \cr &=&& c(\text{Ice}) \cdot m \cdot \Delta(\text{Ice}) \cr & && + c(\text{Water}) \cdot m \cdot \Delta(\text{Water})\cr & && + c(\text{Steam}) \cdot m \cdot \Delta(\text{Steam}) \cr & = && (2.09 \times 13.0 \times 10) \cr & && + (4.182 \times 13.0 \times 100) \cr & &&+ ( 2.01 \times 13.0 \times 10) \cr & = && 5969.6\;\rm J \cr & = && 5.969\; \rm kJ\end{aligned}$ .

<h3>Energy required for step two and four</h3>

The equations for the energy of fusion and energy of vaporization are quite similar:

$E(\text{Fusion}) = n \cdot \Delta H_\text{Fusion}$ .

$E(\text{Vaporization}) = n \cdot \Delta H_\text{Vaporization}$ .

where $n$ is the number of moles of the substance.

Look up the relative atomic mass of oxygen and hydrogen from a modern periodic table:

H: 1.008,
O: 15.999.

Hence the molar mass of water:

$M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g \cdot mol^{-1}$ .

Number of moles of $\rm H_2O$ molecules in $\rm 13.0\; g$ :

$\displaystyle n = \frac{m}{M} \approx 0.721621\; \rm mol$ .

$\begin{aligned}& &&\text{Energy required for phase changes} \cr &=&& n \cdot \Delta H_\text{Fusion} \cr & &&+n \cdot \Delta H_\text{Vaporization} \cr & = &&0.721621 \times 6.02 + 0.721621 \times 40.7 \cr & = &&33.7\; \rm kJ \end{aligned}$

<h3>Energy required for all five steps, combined</h3>

$5.969\; \rm kJ + 33.7\; \rm kJ \approx 39.7\; \rm kJ$ .

8 0

3 years ago

What are the half-reactions for a galvanic cell with aluminum and gold<br> electrodes?

lawyer [7]

Answer:

A

Explanation:

In a galvanic cell, energy is produced by spontaneous chemical processes.

The cathode and anode of this cell will depend on the relative position of the two metals in the electrochemical series.

Aluminium is higher in the electrochemical series so aluminium will be the anode. Silver is lower in the electrochemical series so silver will be the cathode.

Recall that oxidation (electron loss) occurs at the anode while reduction (electron gain) occurs at the cathode.

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3 years ago

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H2(g) + CO2(g) + 10 kcal H2O(g) + CO(g)

EleoNora [17]

<u>Answer:</u> The equilibrium constant for the given reaction is 0.8

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of the products raised to the power its stoichiometric coefficients to the concentration of reactants raised to power its stoichiometric coefficient. It is represented as $K_c$

For the general equation:

$aA+bB\rightleftharpoons cC+dD$

The equilibrium constant is represented as:

$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical equation:

$H_2(g)+CO_2(g)+10 kcal\rightleftharpoons H_2O(g)+CO(g)$

$K_c$ for this equation is given by:

$K_c=\frac{[H_2O][CO]}{[H_2][CO_2]}$

Concentration at equilibrium of

$H_2=2mol/L\\H_2O=4mol/L\\CO_2=5mol/L\\CO=4mol/L$

Putting values in above equation, we get:

$K_c=\frac{4\times 4}{2\times 5}\\K_c=0.8$

Hence, the equilibrium constant for the given chemical reaction is 0.8

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3 years ago

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How much energy is released if a sample loses 0.025 kg mass through<br> radioactive decay?

allochka39001 [22]

Answer:

C

Explanation:

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Describe the concept of using radioactive elements to establish an age for a fossil. Be sure to include how the parent and daugh

Nadusha1986 [10]

Answer:

The nuclear decay of radioactive elements is a process that is a useful tool for determining the absolute age of fossils and rocks. It is used as a clock, in which daughter elements or isotopes converted from parent isotopes by decaying at a particular time.

Radioactive decay rates are constant and do not change over time. It is measured in half-life. A half-life is a time it takes half of a parent isotope to decay and converted into a stable daughter isotope. How many parent isotopes and daughter isotopes present in the fossil or their abundance can help in determining the age of fossil or rock.

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3 years ago