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3 years ago

What is the conjugate base in the following equation: HBr + H2 O \longrightarrow ⟶ H3 O+ + Br-

Chemistry

1 answer:

kolbaska11 [484]3 years ago

7 0

Answer:

$\rm Br^{-}$ would be the conjugate base in this reaction according to the Bronsted-Lowry acid-and-base theory.

Explanation:

Among the reactants, the Bronsted-Lowry acid is the species that supplies protons $\rm H^+$ . The Bronsted-Lowry base would be the species that accept the protons.

On the product side of the equation, the BL-acid donates a proton $\rm H^+$ to produce the conjugate base. The BL-base accepts a proton $\rm H^+$ to produce the conjugate acid.

In this case, each $\rm HBr$ molecule donates a proton to form one $\rm Br^{-}$ ion. Hence,

Each $\rm H_2O$ molecule would accept a proton to produce an $\rm H_3O^{+}$ ion. Hence,

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Answer:

C is the answer they are looking for but clearly the truth is D

Explanation:

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3 years ago

At what time will the pressure of SO₂Cl₂ decline to 0.50 its initial value? Express your answer using two significant figures.

QveST [7]

Answer: The time is 0.69/k seconds

Explanation:

The following integrated first order rate law

ln[SO₂Cl₂] - ln[SO₂Cl₂]₀ = - k×t

where

[SO₂Cl₂] concentration at time t,

[SO₂Cl₂]₀ initial concentration,

k rate constant

Therefore, the time elapsed after a certain concentration variation is given by:

$t=\frac{ln[SO_{2}Cl_{2}]_{0} - ln[SO_{2}Cl_{2}]}{k}=\frac{ln\frac{[SO_{2}Cl_{2}]_{0}}{[SO_{2}Cl_{2}]} }{k}$

We could assume that SO₂Cl₂ behaves as a ideal gas mixture so partial pressure is proportional to concentration:

$p_{(SO_{2}Cl_{2})}V = n_{(SO_{2}Cl_{2})}RT$

$[SO_{2}Cl_{2}]= \frac{n_{(SO_{2}Cl_{2})}}{V}}=\frac{p_{(SO_{2}Cl_{2})}}{RT}}$

In conclusion,

t = ln( p(SO₂Cl₂)₀/p(SO₂Cl₂) )/k

$t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{p_{(SO_{2}Cl_{2})}} }{k}$

for

$p_{(SO_{2}Cl_{2})}=0.5p_{(SO_{2}Cl_{2})}_{0}$

$t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{0.5p_{(SO_{2}Cl_{2})_{0}}} }{k}$

$t=\frac{ln\frac{1}{0.5} }{k}$

$t=\frac{ln(2)}{k}$

$t=\frac{0.69}{k}}$

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3 years ago

Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) â†’ 2Al2O3 (s) In

jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

$\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol$

For oxygen gas:

$\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol$

The chemical equation for the reaction of titanium and chlorine gas follows:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = $\frac{3}{4}\times 0.092=0.069mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of $Al_2O_3$

So, 0.092 moles of aluminium will produce = $\frac{2}{4}\times 0.092=0.046moles$ of $Al_2O_3$

Now, calculating the mass of aluminium oxide:

$\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g$

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

$\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%$

Hence, the percent yield of the reaction is 74 %

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