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3 years ago

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Hydrogen bonds within liquid water are attractions between protons and hydroxide ions. are dipole-dipole attractions. are ion-in

duced dipole attractions. are attractions between protons and oxygen nuclei. are attractions between two hydrogen atoms.

1 answer:

barxatty [35]3 years ago

6 0

Answer:

true because the bonds cannot be broken down

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What is the main side reaction that competes with elimination when a primary alkyl halide is treated with alcoholic potassium hy

Elanso [62]

Answer:

The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.

Explanation:

The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.

A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.

Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.

SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.

7 0

3 years ago

how many grams of fertilizer , for potassium sulfate are there in 42.3 mole of potassium sulfate? step by step

Over [174]

Answer:

The mass of 42.3 moles of potassium sulfate is 7,371.1557 grams

Explanation:

Potassium sulfate K₂SO₄ which is also known as sulphate of potash is a water soluble common component of fertilizer

The molar mass of potassium sulfate, M = 174.259 g/mol

The given number of moles of potassium sulfate, n = 42.3 moles

The mass, 'm', of a given number of moles of potassium sulfate, 'n', is given as follows;

m = n × M

Therefore, we have;

The mass, 'm', of 42.3 moles of potassium sulfate is found by plugging in the values for 'M', and 'n', in the above equation as follows;

m = 42.3 moles × 174.259 g/mol = 7,371.1557 grams

The mass of 42.3 moles of potassium sulfate, m = 7,371.1557 grams.

4 0

3 years ago

What type of atom do you have with p47 and n61

Dmitriy789 [7]

Silver. It is number 47 on the periodic table meaning it has 47 protons. It’s atomic mass is 108. 47+61=108 so it is definitely silver

8 0

3 years ago

In the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. Using the equat

artcher [175]

Answer:do you still need help ?

Explanation:

5 0

3 years ago

Read 2 more answers

how many grams of ammonium carbonate are needed to decompose in order to produce 6.52 g of carbon dioxide?

defon

Answer:

<u>= 14.24g of </u> $(NH4)_{2}CO_{3}$ <u> is required.</u>

Explanation:

Reaction equation:

$(NH4)_{2}CO_{3}$ → $2NH_{3} + CO_{2} + H_{2} O$

Mole ratio of ammonium carbonate to carbon dioxide is 1:1

1 mole of CO2 - 44g

?? mole of CO2 - 6.52g

= 6.52/44 = 0.148 moles was produced from this experiment.

Therefore, if 1 mole of $(NH4)_{2}CO_{3}$ - 96.09 g

0.148 mol of $(NH4)_{2}CO_{3}$ -- ?? g

=0.148 × 96.09

<u>= 14.24g of </u> $(NH4)_{2}CO_{3}$ <u> is required.</u>

5 0

3 years ago