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earnstyle [38]
2 years ago
6

What type of molecule is acetylacetone?

Chemistry
1 answer:
Vlada [557]2 years ago
6 0

Answer:

c. ketone

Explanation:

the answer is ketone

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Explain how you would use the IR spectra to characterize ferrocene, acetyl ferrocene and diacetyl ferrocene.
motikmotik

Answer:

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Explanation:

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3 years ago
Identify whether the following example is qualitative or quantitative data.
OLga [1]
I think that it is qualitative data
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18) What is the following number representing: 10m/s West *
ddd [48]

Speed

Explanation:

The distance travelled by a body per unit time

3 0
3 years ago
Hydrogen peroxide can decompose to water and oxygen by the following reaction
mestny [16]
To begin calculating, there is one thing you need to remember :1 mole of H2O2=34.0148 g
Then we have 5.00g of H2O2=5/34.0148=0.146995
As you know decomposition of 2moles now has prodused <span>196kj
So, </span><span>q is made due </span>0.146995 moles of H2O2=(-196/2)*0.146995=-14.40551Kj
I'm sure it will help.
6 0
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]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

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

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

  • <u>For a:</u>

The given chemical equation follows:

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

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

<u>Reduction half reaction:</u>   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

  • <u>For b:</u>

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.)

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

<u>Reduction half reaction:</u>   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
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