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sveticcg [70]
2 years ago
9

Show all work:

Chemistry
1 answer:
Serggg [28]2 years ago
5 0

The equation of the reaction is;

Ca(OH)2(aq) + 2HCl(aq) ------> CaCl2(aq) + 2H2O(l)

<h3>What is neutralization?</h3>

The term neutralization has to do with the process in which an acid reacts with a base to form salt and water only. A neutralization reaction is known by the fact that water is obtained by the reaction of hydrogen and hydroxide ions.

The equation of the reaction is;

Ca(OH)2(aq) + 2HCl(aq) ------> CaCl2(aq) + 2H2O(l)

Learn more about neutralization:brainly.com/question/27891712

#SPJ1

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State general trend for metal properties as you go left to right across a period
qwelly [4]

Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements.

Electronegativity Trends

Electronegativity can be understood as a chemical property describing an atom's ability to attract and bind with electrons. Because electronegativity is a qualitative property, there is no standardized method for calculating electronegativity. However, the most common scale for quantifying electronegativity is the Pauling scale (Table A2), named after the chemist Linus Pauling. The numbers assigned by the Pauling scale are dimensionless due to the qualitative nature of electronegativity. Electronegativity values for each element can be found on certain periodic tables. An example is provided below.


From left to right across a period of elements, electronegativity increases. If the valence shell of an atom is less than half full, it requires less energy to lose an electron than to gain one. Conversely, if the valence shell is more than half full, it is easier to pull an electron into the valence shell than to donate one.

From top to bottom down a group, electronegativity decreases. This is because atomic number increases down a group, and thus there is an increased distance between the valence electrons and nucleus, or a greater atomic radius.

Important exceptions of the above rules include the noble gases, lanthanides, and actinides. The noble gases possess a complete valence shell and do not usually attract electrons. The lanthanides and actinides possess more complicated chemistry that does not generally follow any trends. Therefore, noble gases, lanthanides, and actinides do not have electronegativity values.

As for the transition metals, although they have electronegativity values, there is little variance among them across the period and up and down a group. This is because their metallic properties affect their ability to attract electrons as easily as the other elements.

According to these two general trends, the most electronegative element is fluorine, with 3.98 Pauling units.



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4 years ago
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frozen [14]

Answer:

Na 1s2 2s2 2p6 3s1

Ar 1s2 2s2 2p6 3s2 3p6

Explanation:

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Which of the following is an immune system disorder ?
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Read 2 more answers
Flag this question question 8 10 pts use the δh°f and δh°rxn information provided to calculate δh°f for if: δh°f (kj/mol) if7(g)
GarryVolchara [31]

\Delta H\textdegree{}_f(\text{IF} \; (g)} = -95 \;  \text{kJ} \cdot \text{mol}^{-1}

Explanation

\text{IF}_7 \; (g) + \text{I}_2 \; (s) \to \text{IF}_5 \; (g) + 2\; \text{IF} \; (g)

  • \Delta H\textdegree{}_\text{rxn} = -89\; \text{kJ} \cdot \text{mol}^{-1}
  • \Delta H\textdegree{}_f (\text{IF}_7 \; (g) ) = -941 \; \text{kJ} \cdot \text{mol}^{-1} (from the question)
  • \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) = -840 \; \text{kJ} \cdot \text{mol}^{-1} (from the question)
  • As an the most stable allotrope under standard conditions, \Delta H\textdegree{}_f (\text{I}_2) = 0\; \text{kJ} \cdot \text{mol}^{-1}

By definition,

\Delta H\textdegree{}_\text{rxn} = \Delta H\textdegree{}_f (\text{all products})  - \Delta H\textdegree{}_f (\text{all reactants})

\Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + 2 \;  \Delta H\textdegree{}_f (\text{IF} \; (g) ) - \Delta H\textdegree{}_f (\text{IF}_7 \; (g) )  - \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) \\  =  \Delta H\textdegree{}_{\text{rxn}}

\begin{array}{ccc} \Delta H\textdegree{}_f  (\text{IF} \; (g) )& = & 1/2\; ( \Delta H\textdegree{}_{\text{rxn}} - \Delta H\textdegree{}_f (\text{IF}_5 \; (g) ) + \Delta H\textdegree{}_f (\text{IF}_7 \; (g) )  + \Delta H\textdegree{}_f (\text{I}_2 \; (s) ) )\\ & = & 1/2 \; (-89 - (-840) + (-941))}\\ & = & - 95 \; \text{kJ} \cdot \text{mol}^{-1} \end{array}

Note, that iodine on the reactant side is stated as a gas in the equation given in the question whereas under standard conditions it is expected to be under the solid state; the \Delta H\textdegree{} _f given in the question seemingly corresponds to the one in which the reactant iodine exists as a solid rather than as a gas. Evaluating the last expression using data from an external source

\Delta H\textdegree{}_f (\text{I}_2 \; (g) ) = \Delta H\textdegree{}_f(\text{I}_2 \; (s)) + \Delta H\textdegree{}_{\text{sublimation}}(\text{I}_2) = 62.42 \;  \text{kJ} \cdot \text{mol}^{-1} (Cox, Wagman, et al., 1984)

... yields \Delta H\textdegree{}_f  (\text{IF} \; (g) )  \approx -64 \; \text{kJ}\cdot \text{mol}^{-1}, which deviates significantly from the experimental value of -94.76    \; \text{kJ}\cdot \text{mol}^{-1} (Chase, 1998.) It is thus assumed that the \Delta H\textdegree{}_\text{rxn} value provided requires a reaction with \text{I}_2 \; (s) rather than \text{I}_2 \; (g) as a reactant.

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Angelina_Jolie [31]

Answer:

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Explanation:

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