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2 years ago

What is the Brønsted-Lowry base in this reaction: NH2−+CH3OH→NH3+CH3O−?(1 point)

Chemistry

1 answer:

Viefleur [7K]2 years ago

7 0

The Brønsted-Lowry base in the given reaction is NH₂⁻. The correct option is the fourth option NH2−

To determine which is the Brønsted-Lowry base in the given reaction:

NH2−+CH3OH→NH3+CH3O−

First, we will write the equation for the reaction properly

The equation is:

NH₂⁻ + CH₃OH → NH₃ + CH₃O⁻

Now, to determine which among the species in the above reaction is the Brønsted-Lowry base, we will start by defining what a Brønsted-Lowry base is.

A Brønsted-Lowry base is any species that is capable of accepting a proton, which requires a lone pair of electrons to bond to the H⁺.

In simple terms, a Brønsted-Lowry base is a proton acceptor.

In the above reaction, NH₂⁻ is the species that is capable of accepting a proton and it has a lone pair of electrons to bond to the H⁺.

∴ NH₂⁻ is the Brønsted-Lowry base in the reaction

Hence, the Brønsted-Lowry base in the given reaction is NH₂⁻. The correct option is the fourth option NH2−

Learn more here: brainly.com/question/13017688

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35mL of acid with an unknown concentration is titrated to completion using 63mL of 3.0 MNaOH.

siniylev [52]

Answer:

5.4 M.

Explanation:

At complete neutralization: It is known that the no. of millimoles of acid equal that of the base.

(MV)acid = (MV)NaOH

M of acid = ??? M, V of acid = 35.0 mL.

M of NaOH = 3.0 M, V of NaOH = 63.0 mL.

∴ M of acid = (MV)NaOH / (V)acid = (3.0 M)(63.0 mL)/(35.0 mL) = 5.4 M.

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3 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

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