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3 years ago

Which is a spectator ion in the reaction between KOH and CoBr2? CoBr2(aq) Co2+(aq) K+(aq) OH– (aq)?

2 answers:

8 0

2KOH(aq) + CoBr₂(aq) → 2KBr(aq) + Co(OH)₂(s)

2K⁺ + 2OH⁻ + Co²⁺ + 2Br⁻ → 2K⁺ + 2Br⁻ + Co(OH)₂

K⁺ and Br⁻ are spectator ions

grandymaker [24]3 years ago

4 0

Answer:

K+ and Br-

Explanation:

A spectator ion is an ion which remains in the same form in both the reactants and product side of a chemical reaction i.e. it does not take part in the reaction.

The reaction between KOH and CoBr2 can be represented as:

2KOH (aq) + CoBr2(aq) → 2KBr(aq) + Co(OH)2(s)

In the above equation the three aqueous solutions, KOH, CoBr2 and KBr will ionize into respective ions unlike Co(OH)2 precipitate. The Total ionic equation can be depicted as:

2K+ + 2OH- + Co2+ + 2Br- → 2K+ + 2Br- + Co(OH)2

The K+ and Br- ions remain unchanged and can be cancelled out giving the net ionic equation:

2OH- + Co2+ → Co(OH)2

Thus, K+ and Br- are spectator ions

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A 1.67-g sample of solid silver reacted in excess chlorine gas to give a2.21-g sample of pure solid Agcl.The heat given off in t

kotegsom [21]

Given:

Mass of Ag = 1.67 g

Mass of Cl = 2.21 g

Heat evolved = 1.96 kJ

To determine:

The enthalpy of formation of AgCl(s)

Explanation:

The reaction is:

2Ag(s) + Cl2(g) → 2AgCl(s)

Calculate the moles of Ag and Cl from the given masses

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3 years ago

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A neutral pH level is _____. 5 7 8

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Neutral pH is level 7

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3 years ago

A critical reaction in the production of energy in biological systems is the hydrolysis of adenosine triphosphate (ATP) to adeno

Archy [21]

Answer:

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

Explanation:

As we can see, we have been a particular reaction and Energy values as well.

ΔG° of reaction = -30.5 kJ/mol

Temperature = 37°C.

And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:

The first step is to calculate the equilibrium constant for the reaction:

Equilibrium Constant K = $\frac{[HPO4-2] x [ADP]}{ATP}$

And we have values given for these quantities in the biological cell:

[HP04-2] = 2.1 x $10^{-3}$ M

[ATP] = 1.2 x $10^{-2}$ M

[ADP] = 8.4 x $10^{-3}$ M

Let's plug in these values in the above equation for equilibrium constant:

K = $\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }$

K = 1.47 x $10^{-3}$ M

Now, we have to calculate the ΔG° of reaction for the biological cell:

But first we have to convert the temperature in Kelvin scale.

Temp = 37°C

Temp = 37 + 273

Temp = 310 K

ΔG° of reaction = (-30.5 $10^{3}$ ) + (8.314)x (310K)xln(0.00147)

Where 8.314 = value of Gas Constant

ΔG° of reaction = (-30.5 x $10^{3}$ ) + (-16810.68)

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3 years ago