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3 years ago

What is a precipitate?

Chemistry

2 answers:

zloy xaker [14]3 years ago

8 0

Answer:

Hi there!

A precipitate is the creation of a solid from a chemical reaction of liquids

Hope this helps!

qwelly [4]3 years ago

8 0

Answer:

it's a cause(an event or situation, typically one that's is bad or undesirable) to happen suddenly, unexpected, or prematurely.

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_H2+O2=_H2O balance help please

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Both are 2
2H2+O2=2H2O

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3 years ago

What is a promblem based on charles' law?

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Answer:

You will never know the exact volume with charles law

Explanation:

Doubling the temperature of gas doubles its volume, so long as the pressure and quantity of the gas are unchanged.

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3 years ago

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What is the number of moles in 3.0 X 10^24 atoms of Carbon

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Answer:the number of moles represented by 3.0 x 10^24 atoms of Ag is 0.500mol 0.500 m o l .

Explanation:

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1 year ago

Which model of the atom is the most accurate

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Quantum Mechanical model (as per my knowledge)

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3 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago