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3 years ago

Draw the products for the proton transfer reaction between sodium hydride and ethanol

1 answer:

5 0

Sodium hydride has the formula NaH where we have a sodium ion, Na⁺ and a hydride ion, H⁻. Hydride is an incredibly powerful base. While it is capable of acting as a nucleophile, if there is an acidic proton in a molecule, the hydride will deprotonate the molecule and grab the most acidic proton.

The pka of H⁻ is 35. The pka of ethanol is 16. The species with the larger pka is the better base and is capable of deprotonating the species with the smaller pka. Therefore, the hydride will deprotonate the acidic -OH proton of the alcohol in the following reaction:

CH₃CH₂OH + NaH → CH₃CH₂O⁻Na⁺ + H₂

The result of the reaction is the hydride deprotonates the proton of the alcohol and forms the alkoxide, which is a sodium salt. This reaction also leads to the formation of H₂ gas which ensures that this reaction is not reversible as the H₂ leaves the reaction mixture upon formation.

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Which of an atom's electrons are inrolved in chemical reactions?

Scrat [10]

Answer:

Electrons at the outermost energy level of an atom are called valence electrons. They determine many of the properties of an element. That's because these electrons are involved in chemical reactions with other atoms. Shared electrons bind atoms together to form chemical compounds.

Explanation:

8 0

2 years ago

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Relative dating determines the relative age of fossils according to

KiRa [710]

The answer Is C.the layer of rock they're found in

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2 years ago

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

<u>Answer:</u> The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

<u>Initial:</u> 0.025 0.025

<u>At eqllm:</u> 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

2 years ago

Find [cu2+] in a solution saturated with cu4(oh)6(so4) if [oh − ] is fixed at 2.3 ✕ 10−6 m. note that cu4(oh)6(so4) gives 1 mol

ki77a [65]

<span>

</span> $Cu_4(OH)_6(SO_4)$
<span>
You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3

So, for 2.3 ✕ 10−6 m OH-
[Cu2+] = $\frac{2}{3} \times 2.3 \times 10^{-6}$

$= 1.53 \times 10^{-6} 
$

6 0

3 years ago

A eraser has a mass of 4g and a volume of 2cm3 what is it’s density

sladkih [1.3K]

Answer:

<h3>The answer is 2.0 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 4 g

volume = 2 cm³

We have

$density = \frac{4}{2} \\$

We have the final answer as

Hope this helps you

4 0

3 years ago