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3 years ago

Sexual reproduction _____. (Choose all that apply)

Chemistry

2 answers:

Sergio039 [100]3 years ago

5 0

2 and 5 would be the ones that apply

Murrr4er [49]3 years ago

4 0

2 and 5 are features of sexual reproduction.

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How to convert n butylbenzene to benzoic acid

Pavlova-9 [17]

Answer:

When an alkyl benzene is heated with strong oxidizing asgents like acidic or alkline KMnO4

or acidified K2Cr2O7

, etc. gives aromatic carboxyllic acid. The alkyl side chain gets oxidised to −COOH

group irrespective of the size of the chain.

Explanation:

4 0

2 years ago

What is the oxidation number of chlorine

sergiy2304 [10]

The oxidation number of chlorine in the Cl- ion is -1.

3 0

2 years ago

Which of the following statements is not an accurate description of a factor that contributes to the ordering of the spectrochem

MArishka [77]

Answer:

Higher oxidation state metals form stronger bong with ligands

Explanation:

Ligand strength are based on oxidation number, group and its properties

5 0

3 years ago

The balanced redox reactions for the sequential reduction of vanadium are given below.

Minchanka [31]

Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

$2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)$ $2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)$

$2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)$

On adding above equations, we get the following combined equation.

$2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)$

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

$12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}$

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

$0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn$

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

$0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn$

We need 0.0392 grams of Zn metal to completely reduce vanadium.

6 0

3 years ago

In ionic bonds, valence electrons are

Serjik [45]

Answer:

Valence Electrons are transferred/exchanged

6 0

2 years ago