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3 years ago

CHEMISTRY. PLEASE HELP!!

Chemistry

2 answers:

Zina [86]3 years ago

6 0

Answer:

B: It is positive when the bonds of the product store more energy than those of the reactants.

Explanation:

took the test

bixtya [17]3 years ago

3 0

Answer:

The answer is B on edge

Explanation:

Here are my notes on this section for anyone that needs them

Enthalpy and State Function

Bonds contain potential energy. Breaking and forming bonds involves energy. Reactants and products contain energy. Enthalpy (H) is a measure of heat and internal energy in a system.

A state function is a quantity whose change in magnitude during a process depends only on the beginning and end points the process, not the path taken between them. Enthalpy change during reaction depends only on the identity of reactants and products and their initial and finial condition

Enthalpy of Formation

enthalpy of formation (Hf) is the energy absorbed or released when a pure substance forms from elements in their standard states

Units: kJ/mol, kcal/mol

Standard state is the natural state of an element at 1 atm (atmosphere of pressure) and 25 degrees celsius. Hf for a pure element in its standard state is 0 kJ/mol.

H (hydrogen): H2(g)

N (nitrogen): N2(g)

O (oxygen): O2(g)

F (fluorine): F2(g)

Cl (chlorine): Cl2(g)

Br (bromine): Br2(l)

Hg (mercury): Hg(l)

Enthalpy of Reaction

Enthalpy of reaction (Hrxn) is energy absorbed or released during a chemical reaction

Hrxn negative: exothermic reaction

Hrxn positive: endothermic reaction

Hess's Law: Hrxn = Σ(ΔHƒ, products) − Σ(ΔHƒ, reactants)

thermochemical equation: the chemical equation that shows the state of each substance involved and the energy change involved in a reaction

Find the kJ/mol of the product and then subtract the kJ/mol of the reactants.

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3 years ago

A flexible container at an initial volume of 5.120 L contains 8.500 mol of gas.

SOVA2 [1]

Answer:

30.05 mol

Explanation:

solving the proportion

V1 / n1 = V2 / n2

5.120 L 18.10 L

–––––––– = ––––––

8.500 mol x

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3 years ago

2. What ions are present in what ratio in a solution of aqueous calcium chloride?

Alenkasestr [34]

Answer:

$\mathrm{Ca}^{2+} \text { and } \mathrm{Cl} \text { - ions are present in } 1: 2 \text { ratio in a solution of aqueous calcium chloride. }$

Explanation:

Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form $\mathrm{Ca}^{2+}$ , while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two $\mathrm{Cl}^{-}$ ions.

$\text { Equation is as follows: } \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_{2}$

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one $\mathrm{Ca}^{2+}$ and two $\mathrm{Cl}^{-}$ ions: $\mathrm{CaCl}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Ca}^{2+}(\mathrm{ag})+2 \mathrm{Cl}(\mathrm{ag})$

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4 years ago

A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so

omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: $AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s)$ .

Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
Secondly, addition of liquid ammonia would form a precipitate with silver: $AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)$ ; Silver hydroxide at higher temperatures decomposes into black silver oxide: $2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)$ .
Thirdly, we also know we have $Hg_2^{2+}$ in the mixture, since addition of potassium chromate produces a yellow precipitate: $Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s)$ . The latter precipitate is yellow.

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4 years ago

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4 0

4 years ago

Read 2 more answers