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Vikki [24]
4 years ago
7

Naming Binary compounds (ionic)?1) BaCl2 ___ _______ 2) NaF _____________ 3) CuBr ___________ 4) CuBr2 ___________ 5) Ba3P2 ____

_______ 6) FeO ________________ 7) Fe2O3 ______________ 8) MgS __________ 9) Al2O3 ____________ 10) CaI2 ______________ 11) K2S __________ 12) CrCl2 ______ 13) CrCl3 _____ 14) CaO ____________ 15) Ag2O ______________ 16) ZnO _______________ 17) Na2O ____________ 18) BeS ___________ 19) MnO _______ 20) Mn2O3 ______
Chemistry
1 answer:
IgorLugansk [536]4 years ago
3 0

Answer:

1) BaCl2 Barium chloride

2) NaF Sodium fluoride

3) CuBr Copper (I) Bromide

4) CuBr2 Copper (II) bromide

5) Ba3P2 Barium phosphide

6) FeO Ferrous oxide

7) Fe2O3 Ferric oxide

8) MgS Magnesium sulfide

9) Al2O3 Aluminum oxide

10) CaI2 Calcium iodide

11) K2S Potassium sulfide

12) CrCl2 Chromous chloride

13) CrCl3 Chromium (III) chloride

14) CaO Calcium oxide

15) Ag2O Silver (I) oxide

16) ZnO Zinc oxide

17) Na2O Sodium oxide

18) BeS Beryllium sulfide

19) MnO Manganese (II) oxide

20) Mn2O3 Manganese (III) oxide

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(a) ions in a certain volume of .20M NaCl (aq) are represented in the box above on the left. in the box above on the right, draw
mamaluj [8]

Answer:

  • The first picture attached is the diagram that accompanies the question.

  • The<u> second picture attached</u> is the diagram with the answer.

Explanation:

In the box on the left there are 8  Cl⁻ ions and 8 Na⁺ ions.

The dissociaton equation for NaCl(aq) is:

  • NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)

The dissociation equation for CaCl₂ (aq)  is:

  • CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)

A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.

Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half)  and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).

Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.

4 0
3 years ago
What is the molality of a solution of 0.25 moles of sucrose in dissolved in enough water to make 4.00 liters of water?
Grace [21]
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Your tongue has about 3,000 taste buds.<br> True<br> False
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The octet rule is the basis for the stability of the atom. According to this rule, the outermost shell of the atom must contain
liubo4ka [24]

Answer:8

Explanation:

6 0
3 years ago
Methane, a component in natural gas, can be used as a fuel in combustion reactions. What is the value for ΔGnon (in kJ) for the
Oksanka [162]

<u>Answer:</u> The \Delta G for the reaction is -806.86 kJ

<u>Explanation:</u>

We are given:

\Delta H^o_{rxn}=-803kJ=-803000J      (Conversion factor:  1 kJ = 1000)

\Delta S^o_{rxn}=-4.05J/K

Temperature of the reaction = 293 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delat H^o_{rxn}-T\Delta S^o_{rxn}

Putting values in above equation, we get:

\Delta G^o_{rxn}=-803000J-[(293K)\times (-4.05J/K)]=-801813.35J

For the given chemical equation:

CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(g)

The expression for K_c is given as:

K_{c}=\frac{[H_2O]^2[CO_2]}{[CH_4][O_2]^2}

We are given:

[H_2O]=6.41M

[CO_2]=3.83M

[CH_4]=14.51M

[O_2]=9.27M

Putting values in above equation, we get:

K_c=\frac{(6.41)^2\times 3.83}{14.51\times (9.27)^2}

K_c=0.126

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_c

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = -801813.35 J

R = Gas constant = 8.314J/K mol

T = Temperature = 293 K

K_c = equilibrium constant in terms of concentration = 0.126

Putting values in above equation, we get:

\Delta G=-801813.35J+(8.314J/K.mol\times 293K\times \ln(0.126))

\Delta G=-806859.46J=-806.86kJ

Hence, the \Delta G for the reaction is -806.86 kJ

8 0
3 years ago
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