Why is it important to keep your apparatus dry what reaction will occur between the grignard reagent and water?

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Answer 1

Answer: Grignard reagents react with water to create Alkanes. It is important to keep the apparatus dry because grignard reagents react with water

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Choose all the answers that apply.Ionic compounds _____.
-do not dissolve in water
-have high melting points
-have low melting points
-dissolve easily in water
-conduct electricity when melted

Answers

Answer:

high melting points, dissolve easily in water, conduct electricity when melted.

Explanation:

I'm pretty sure

7. A change of state is a(n)_process.
a. Irreversible
b. Reversible

Answers

Answer: b

A change of state is a reversible process.

B that’s the answer

Good luck

The following figure represents the formation of an ionic compound. Substances A and B are initially uncharged, but when mixed electrons are transferred. Using the figure, identify which substance will form the cation and which will form the anion. Provide a brief (one or two sentences) explanation for your response. (Hint: How does losing electrons affect atomic radii?)

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete but some general explanation provides a clear answer to what is been asked in the question.

An ionic/electrovalent compound is a compound whose constituent atoms are joined together by ionic bond. Ionic bond is a bond involving the transfer of valence electron(s) from an atom (to form a positively charged cation) to another atom (to form a negatively charged anion). The atom transferring is usually a metal while the atom receiving is usually a non-metal.

For example (as shown in the attachment), in the formation of NaCl salt, the sodium (Na) transfers the single electron (valence) on it's outermost shell to chlorine (Cl) which ordinarily has 7 electrons on it's outermost shell but becomes 8 after receiving the valence electron from sodium. It should also be noted that Na is a metal while Cl is a non-metal.

Dinitrogen monoxide has a structural formula of NNO and requires resonance structures in order to draw the Lewis structures of the molecule. Based on formal charge distributions, themostsignificant (stable) resonance structure for this molecule exhibits the order of formal charges for the 1st N, the central N, and the O atoms, respectively, as:A. 0,+1,-1
B. -1,+1,0
C. -2,+3,-1
D. 0,0,0

Answers

Three resonance structures contribute to the structure of dinitrogen monoxide.

The resonance structure is invoked when a single structure can not sufficiently explain all the bonding properties of a compound. All the various contributing structures contribute to the final structure of the compound but not all to the same degree.

There are three resonance structures of dinitrogen monoxide. The most stable structure is always the structure that has the formal charges as -1, +1 and zero as shown.

Learn more: brainly.com/question/14283892

Answer:

A. 0, +1, -1

Explanation:

You can draw the lewis structure for NNO 3 ways: With two double bonds N=N=O, with a triple bond between the N and O and single bond between the two N's, or a triple bond between the two N's and a single bond between the N and O.

The goal is to have formal charges that are as small as possible, to have no identical formal charges on adjacent atoms, and to have the most negative formal charge on the most electronegative atom. The most stable structure is the one with the triple bond between the two N's because it gives the formal charges 0, 1, and -1 respectively. Unlike the other two structures, the negative formal charge is correctly placed on O, the most electronegative atom.

What product would you expect from a nucleophilic substitution reaction of (S)-2-bromohexane with acetate ion, CH3CO22? Assume that inversion of configuration occurs, and show the stereochemistry of both the reactant and product.

Answers

Answer:

(R) - hexyl acetate

Explanation:

Hello,

This reacción is a nucleophilic substitution SN2.

The configuration (s), means that the groups around the chiral carbon are organized appose to the clock hands movement. But when the reaction happens, these configurations become an (r) configuration, it means the groups around the chiral carbon organize according to the clock hands movement.

Generally, these reactions are related to nucleophilic species, an example is the ion acetate, a conjugated acid which is a weak nucleophilic, for this reason, the transition state is more energetic, it means, less stable than if the reaction occurs with a strong nucleophilic.

Look the image to compare the two configurations of the reactant and product.

Part D 2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s) Drag the appropriate labels to their respective targets. ResetHelp e−→e Superscript- rightarrow ←e−leftarrow e Superscript- Cathode Cathode Anode Anode II Superscript- ClO−2C l O Subscript 2 Superscript- Request Answer Part E Indicate the half-reaction occurring at Anode. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing

Answers

The half reaction occurring at anode is:

$2I^-(aq)---- > I_2(s)+2e^-$

Half reaction for the cell:

The substance having highest positive potential will always get reduced and will undergo reduction reaction.

Balanced chemical equation:

$2ClO_2(g)+2I^-(aq)----- > 2ClO^(2-)(aq)+I_2(s)$

The half reaction follows:

Oxidation half reaction: $2I^-(aq)---- > I_2(s)+2e^-$ , Reduction potential is 0.53V

Reduction half reaction: $ClO_2(g)+e^----- > ClO_2^-$ ( × 2 ), Oxidation potential is +0.954 V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is :

$2I^-(aq)---- > I_2(s)+2e^-$

Find more information about Reduction potential here:

brainly.com/question/7484965

Answer: The half reaction occurring at anode is $2I^-(aq.)\rightarrow I_2(s)+2e^-$

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

For the given chemical equation:

$2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^(-2)(aq.)+I_2(s)$

The half reaction follows:

Oxidation half reaction: $2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_(I_2/I^-)=0.53V$

Reduction half reaction: $ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_(ClO_2/ClO_2^-)=+0.954V$ ( × 2 )

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is $2I^-(aq.)\rightarrow I_2(s)+2e^-$

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