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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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A compound is found to contain 38.65 % carbon, 16.25 % hydrogen, and 45.09 % nitrogen by mass. what is the empirical formula for

iren [92.7K]

The molar mass of carbon is 12, hydrogen is 1, and nitrogen is 14, hence the ratio are:

C = 38.65 / 12 = 3.22

H = 16.25 / 1 = 16.25

N = 45.09 / 14 = 3.22

Divide the three by the lowest ratio which is 3.22:

C = 3.22 / 3.22 = 1

H = 16.25 / 3.22 = 5

N = 3.22 / 3.22 = 1

So the empirical formula is:

CHN

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When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

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Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

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