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Tju [1.3M]
3 years ago
7

A chemist who studies water samples did a demonstration of how to test for lead in water. She added a clear solution of potassiu

m iodide to a clear solution of lead nitrate. Then a yellow swirling solid formed in the liquid. What is most likely true about the yellow solid?
Chemistry
2 answers:
evablogger [386]3 years ago
8 0

<em>I actually need help myself with question and no one has helped me yet ? </em>

Lerok [7]3 years ago
3 0

Answer:

The swirling yellow solid formed is lead iodide (PbI₂).

Explanation:

  • The reaction of potassium iodide (KI) with lead nitrate (Pb(NO₃)₂) will produce lead iodide (PbI₂) and potassium nitrate (KNO₃) according to the equation:

2KI + Pb(NO₃)₂ → PbI₂↓ + 2KNO₃

  • Lead iodide (PbI₂) is a yellow swirling precipitate that is formed from the reaction.
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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)

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

5 0
3 years ago
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