In the first step, 4-sulfanilic acid reacts with sodium nitrate to form diazonium ion intermediate. Identify the Lewis acid and Lewis base in this reaction.

Answers

Answer 1

Answer:

Answer:

In the formation of diazonium ion intermediate, the 4-sulfanilic acid acts as the Lewis acid, while the sodium nitrite is the Lewis base.

Explanation:

A Lewis acid is by definition an electron pair acceptor (such as the H+ ion, that can accept a pair of non-bonding electrons) and a Lewis base is an electron pair donor (such as the OH- ion, that can donate a pair of non-bonding electrons).

In the formation of diazonium ion intermediate, the 4-sulfanilic acid acts as the Lewis acid(by transference of a lone pair from its nitrogen atom), while the sodium nitrite is the Lewis base.

N.B its sodium nitrite, NaNO2 (which is slightly basic in solution) not nitrate NaNO3 (which is neutral in solution)

Answer 2

Answer:

4-sulfanilic acid acts as Lewis acid, while Sodium nitrite is the Lewis base.

What are Lewis acid and Lewis base?

A Lewis acid is an electron pair acceptor (such as the H+ ion, that can accept a pair of non-bonding electrons) and a Lewis base is an electron pair donor (such as the OH- ion, that can donate a pair of non-bonding electrons).

In the formation of diazonium ion intermediate, the 4-sulfanilic acid acts as the Lewis acid(by transference of a lone pair from its nitrogen atom), while the sodium nitrite is the Lewis base.

Sodium nitrite, NaNO₂ (which is slightly basic in solution) not nitrate NaNO₃ (which is neutral in solution)

Find more information about Lewis acid and base here:

brainly.com/question/15220646

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What is the concentration of hydronium ion ( [H3O+]) in a solution with a PH of _1,3?

Answers

Answer: [H3O+]= 0.05 M

Explanation:

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The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 7.300 g C 6 H 6 7.300 g C6H6 is burned and the heat produced from the burning is added to 5691 g 5691 g of water at 21 ∘ 21 ∘ C, what is the final temperature of the water?

Answers

Answer: The final temperature of the water is $33.85^(o)C$ .

Explanation:

We know that molar mass of $C_(6)H_(6)$ is 78 g/mol. And, the amount of heat produced when 2 mol of $C_(2)H_(6)$ burns is 6542 KJ.

This means that,

$78 * 2$ = 156 g of $C_(2)H_(6)$ burns, heat produced is 6542 kJ.

Therefore, heat produced (Q) by burning 7.3 g of $C_(6)H_(6)$ is as follows.

$(6542 * 7.3 g)/(156 g)$

= 306.13 kJ

or, = 306130 J (as 1 KJ = 1000 J)

For water, mass is given as 5691 g and specific heat capacity of water is 4.186 $J/g^(o)C$ .

So, we will calculate the value of final temperature as follows.

Q = $m * C * (T_(f) - T_(i))$

306130 J = $5691 g * 4.186 J/g^(o)C * (T_(f) - 21)^(o)C$

$(T_(f) - 21)^(o)C = (306130 J)/(23822.53 J/^(o)C)$

$T_(f)$ = 12.85 + 21

= $33.85^(o)C$

Thus, we can conclude that the final temperature of the water is $33.85^(o)C$ .

Consider the combination reaction of samarium metal and oxygen gas. If you start with 33.7 moles of samarium metal, how many moles of oxygen gas would be required to react completely with all of the samarium metal? For this reaction, samarium has a +3 oxidation state within the samarium/oxygen compound.

Answers

Answer:

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Explanation:

$4Sm+3O_2\rightarrow 2Sm_2O_3$

Number of moles samarium metal = 33.7 moles

According to reaction, 4 moles of samarium reacts with 3 moles of oxygen gas.

Then 33.7 moles of samarium will react with:

$(3)/(4)* 33.7 mol=25.275 mol$ of oxygen gas.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Answer:

Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is $\fbox{25.3 \text{ mol}}$ .

Explanation:

A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.

The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:

1. Amount of one reactant required to react completely with the other reactant.

2. Amount of the product that can be produced from the given amount of the reactant.

Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.

The chemical formula for oxygen gas is $\text{O}_(2)$ .

Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is $\text{Sm}_(2)\text{O}_(3)$ .

The chemical equation is as follows:

$\fbox{\text{Sm}+\text{O}_(2) \rightarrow \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.

The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of $\text{Sm}_(2)\text{O}_(3)$ and 3 in front of $\text{O}_(2)$ to balance the oxygen atoms.

$\fbox{\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.

$\fbox{\n4\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.

According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.

Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:

$\text{moles of O}_(2) = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)$ ...... (1)

Step 5: Substitute 33.7 mol for moles of Sm in equation (1).

$\text{Moles of O}_(2) = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)\n\text{Moles of O}_(2)= 25.275 \text{ mol}\n\text{Moles of O}_(2)= 25.3 \text{ mol}$

Note:

Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Some basic concept of chemistry

Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.

A lab group was calculating the speed of a radio car. They measured the distance traveled to be 6 meters and the time to be 3.5 seconds. Then they divided the distance by the time to find the speed. The actual speed was 2.2 m/s. What was their percent error?

Answers

Percent error is a measure of the difference between an observed value and a true value.

Actual Speed (True Value) = 2.2 m/s

Experimental Speed (Calculated Value) = Distance / Time = 6 m / 3.5 s = 1.714 m/s

The formula for calculating percent error is:

Percent Error = ((|Actual Value - Experimental Value|) / |Actual Value|) * 100%

Calculate the absolute difference between the actual speed and the experimental speed:

|2.2 - 1.714| = 0.486

Calculate the absolute value of the actual speed:

|2.2| = 2.2

Percent Error = (0.486 / 2.2) * 100%

= 0.221 * 100%

= 22.1%

The calculated percent error is approximately 22.1%. This means that the lab group's calculated speed of 1.714 m/s is about 22.1% lower than the true speed of 2.2 m/s.

Percent error is a way to quantify the accuracy of experimental measurements. A positive percent error indicates that the experimental value is higher than the true value, while a negative percent error indicates that the experimental value is lower. In this case, since the calculated speed is lower than the true speed, we have a positive percent error.

To know more about actual speed here

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Answer:456

Explanation:

Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0 L buffer containing 0.252 M NaHSO3 and 0.139 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH? Enter your answer numerically to 4 decimal places.

Answers

Answer:

pH = 7.1581

Explanation:

The equilibrium of NaHSO₃ with Na₂SO₃ is:

HSO₃⁻ ⇄ SO₃²⁻ + H⁺

Where K of equilibrium is the Ka2: 6.5x10⁻⁸

HSO₃⁺ reacts with NaOH, thus:

HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

HSO₃⁻: 0.252 moles

SO₃²⁻: 0.139 moles

Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

0.0500L ₓ (1mol /L): 0.050 moles of NaOH.

Thus, final moles of both compounds are:

HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles

SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:

pH = pka + log [SO₃²⁻] / [HSO₃⁻]

Where pKa is - log Ka = 7.187

Replacing:

pH = 7.187 + log [0.189] / [0.202]

pH = 7.1581

A 0.7 ft diameter hole forms in a tank containing butane at 19 atmg and 76 degrees Fahrenheit. Determine the maximum possible mass flow rate through this leak in lb m / min, if the external pressure is 1 atm.