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3 years ago

Please help if you can

Chemistry

1 answer:

White raven [17]3 years ago

3 0

Answer:

i cants bye

Explanation:

byeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

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Which one of the following reactions would produce t-butyl methyl ether in high yield?

swat32

Answer:

D) sodium t-butoxide + bromomethane

Explanation:

The alkoxide ion is a strong nucleophile, that unlike alcohols, will react with primary alkyl halides to form ether. This general reaction is known as <em>the Williamson synthesis</em>, and is a SN₂ displacement. The alkyl halide must be primary so the back side attack is not hindered, and the alkoxide ion must be formed with the most hindered group.

The mechanism can be seen in the attachment.

7 0

4 years ago

A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric ac

UkoKoshka [18]

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

$pH = pKa + log(\frac{[KF]}{[HF]})$

$pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04$

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq) + F⁻(aq) ⇄ HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

$\eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol$

$\eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles$

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

$\eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles$

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq) (1)

The number of moles of HF after the reaction of KF with HCl is:

$\eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles$

And the concentration of HF after the reaction of KF with HCl is is:

$C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L$

Now, from the equilibrium of equation (1) we have:

$Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]}$

$Ka = \frac{x^{2}}{0.531 - x}$ (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

$pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73$

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

8 0

4 years ago

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or io

Dmitrij [34]

Answer :

(a) The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

The spectator ions are, $NH_4^{+}\text{ and }SO_4^{2-}$

(b) The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

(c) The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation will be,

$Cr_2(SO_4)_3(aq)+3(NH_4)_2CO_3(aq)\rightarrow 3(NH_4)_2SO_4(aq)+Cr_2(CO_3)_3(s)$

The ionic equation in separated aqueous solution will be,

$2Cr^{2+}(aq)+3SO_4^{2-}(aq)+6NH_4^{+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)+6NH_4^{+}(aq)+3SO_4^{2-}(aq)$

In this equation, $NH_4^{+}\text{ and }SO_4^{2-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

(b) The given balanced ionic equation will be,

$Ba(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow 2KNO_3(aq)+BaSO_4(s)$

The ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

(c) The given balanced ionic equation will be,

$Fe(NO_3)_2(aq)+2KOH(aq)\rightarrow 2KNO_3(aq)+Fe(OH)_2(s)$

The ionic equation in separated aqueous solution will be,

$Fe^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

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4 years ago

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shusha [124]

It would be D.the food chain

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3 years ago

The ________ is the energy difference between reactants and products in a chemical reaction. heat of reaction transition energy

bearhunter [10]

The answer is "heat of the reaction"

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