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qwelly [4]
3 years ago
14

Identify the spectator ions in the following complete ionic equation.

Chemistry
1 answer:
insens350 [35]3 years ago
3 0

Answer:

2I⁻  and  2Na⁺  are spectator ions.

Explanation:

Ionic equation:

Ba⁺²(aq)  + 2I⁻(aq) + 2Na⁺(aq)  + SO₄²⁻ (aq)   →   BaSO₄(s)  +  2I⁻ (aq) + 2Na⁺(aq)

Net ionic equation:

Ba⁺²(aq)  + SO₄²⁻ (aq)   →   BaSO₄ (s)  

The I⁻ and Na⁺  are spectator ions that's why these are not written in net ionic equation. The BaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

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We calculate the molar concentration [Cl⁻]  using stoichiometry. MnCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MnCl2 that dissolves.
MnCl2(s) --> Mn+(aq) + 2 Cl⁻(aq)
            [Cl⁻] = 0.68 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.4 M
The answer to this question is [Cl⁻] = 1.4 M

5 0
3 years ago
Which statement about Niels Bohr's atomic model is true?
cupoosta [38]

Answer:

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Explanation:

6 0
3 years ago
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Copper atoms are used to produce pennies. The copper atoms in pennies share many electrons. Pennies contain
kaheart [24]

Answer:

Metallic bonds

Explanation:

Metallic bonds joins atoms of metals and atoms of alloys together. The copper used in making pennies is a metallic substance so it contains metallic bonds.

  • The formation of this bond type is predicated on the large atomic radius, low ionization energy and large number of electrons in the valence shell.
  • The bond is an attraction between the positive nuclei of all closely packed atoms in the lattice and the electron cloud.
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This way the bond in pennies are metallic in nature.

5 0
3 years ago
2. Hydrogen gas at a temperature of 22.0°C that is confined in a 5.00L cylinder exerts a pressure of 4.20atm. If the gas is rele
Umnica [9.8K]

Answer: n∗R=22+273.15/4.2∗5n

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Explanation:

4 0
3 years ago
A gaseous mixture at 265K and 1.0 atm contains 25 O2 60 N2 and 15 CO2 mole basis The velocities of the components are 0.084 cm s
Free_Kalibri [48]

Answer:

The molar average velocity is 0.0588 cm/s

The N₂ diffusion velocity relative to the mole average velocity is -0.1428 cm/s

The molar diffusional flux of N₂ is -3.9x10⁻³

Explanation:

Given data:

T = temperature = 265 K

O₂ = 25%

N₂ = 60%

CO₂ = 15%

vO₂ = -0.084 cm/s

vN₂ = 0.12 cm/s

vCO₂ = 0.052 cm/s

The molar average velocity is equal:

v_{av} =(0.25*(-0.084))+(0.6*0.12)+(0.15*0.052)=0.0588cm/s

The N₂ diffusion velocity relative to the molar average velocity is:

v_{i} -v_{av} =-0.084-0.0588=-0.1428cm/s

The molar diffusional flux of N₂ is:

N_{N_{2} } =\frac{P}{RT} y_{A} (v_{i} -v_{av} )=-3.9x10^{-3}

8 0
4 years ago
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