Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H2O). Suppose 5.7 g of hydrobromic acid is mixed with 0.980 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answers

Answer 1

Answer:

The maximum mass of water that could be produced by the chemical reaction=0.441g

Explanation:

We are given that

Given mass of HBr=5.7 g

Given mass of sodium hydroxide=0.980 g

Molar mass of HBr=80.9 g/ Mole

Molar mass of NaOH=40 g/mole

Molar mass of H2O=18 g/mole

Reaction

[tex]HBr+NaOH\rightarrow H_2O+NaBr[/tex]

Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]

Using the formula

Number of moles of HBr=[tex]\frac{5.7}{80.9}=0.0705 moles[/tex]

Number of moles of NaOH=[tex]\frac{0.980}{40}=0.0245moles[/tex]

Hydrogen bromide is in a great excess and the amount of water produced.

Therefore,

Number of moles of water, n(H2O)=Number of moles of NaOH=0.0245moles

Now,

Mass of water=[tex]n(H_2O)\times Molar\;mass\;of\;water[/tex]

Mass of water=[tex]0.0245moles\times 18=0.441g[/tex]

Hence, the maximum mass of water that could be produced by the chemical reaction=0.441g


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Answer:

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Answer:

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Answers

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Answer:

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Answer:

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Answer:

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Answers

Answer:

Consider the following equilibrium:

2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.

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Explanation:

Given,

[tex]Kc=1.08 * 10^7[/tex]

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=-1.

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[tex]Kp=1.08 * 10^7 / 8.314 J.K6-1.mol^-^1 x 973 K\\Kp=1335.06[/tex]

The answer is Kp=1335.06

The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].

Explanation:

The relation between [tex]K_p \& K_c[/tex] is given by:

[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]

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[tex]K_p[/tex] = The equilibrium constant of reaction in terms of partial pressure

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T = The temperature of the equilibrium

[tex]n_g[/tex]= Change in gaseus moles

Given:

An equilibrium reaction, 700°C:

[tex]2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g),K_c=1.08\times 10^7[/tex]

To find:

The equilibrium constant in terms of partial pressure, [tex]K_p[/tex].

Solution:

The equilibrium constant of reaction in terms of concentration= [tex]K_c[/tex]

[tex]K_c=1.08\times 10^7[/tex]

The equilibrium constant of reaction in terms of partial pressure =[tex]K_p=?[/tex]

The gaseous moles of reactant side = [tex]n_r= 3[/tex]

The gaseous moles of product side = [tex]n_p= 2[/tex]

The temperature at which equilibrium is given = T

[tex]T = 700^oC+273.15 K=973.15K[/tex]

The change in gaseous mole  = [tex]n_g=n_p-n_r=2-3 = -1[/tex]

[tex]K_p=1.08\times 10^7\times (0.0821 atm L/mol K\times 973.15 K)^{-1}\\K_p=1.35\times 10^5[/tex]

The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].

Learn more equilibrium constants here:

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Answer:

True

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Which of the following statements correctly explains why bromination reactions are more selective than chlorination reactions.
a. bromine radical is less stable than chlorine radical, so it is more reactive and less choosy
b. bromine radical is more stable than chlorine radical, so it is more reactive and less choosy
c. bromine radical is more stable than chlorine radical, so it is less reactive and more choosy
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Answer: A bromine radical is more stable than chlorine radical, so it is less reactive and more choosy.

Explanation:

A chlorine atom being more electronegative in nature is able to attract a hydrogen atom more readily towards itself as compared to a bromine atom.

Since bromine is less electronegative in nature so bromine will be more selective as a hydrogen abstracting agent. As a result, bromine radical is more stable in nature than chlorine radical.

Thus, we can conclude that bromine radical is more stable than chlorine radical, so it is less reactive and more choosy.

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Answer:

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Yes, each chemical substance has their own chemical composition so a formula is used to show this.

Yes.

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a sample of salt has 1.74 moles of sodium chloride. how many formula units of the ionic compound are in the sample?

Answers

Answer:

A sample of salt has 1.74 moles of sodium chloride. how many formula units of the ionic compound are in the sample?

Explanation:

Given, 1.74 moles of NaCl.

Since one mole of NaCl consists of --- [tex]6.023 * 10^2^3[/tex] formula units.

Then, 1.74mol of NaCl contains how many formula units of NaCl?

[tex]1.74 mol x \frac{6.023x10^2^3}{1 mol} \\=10.5x10^2^3[/tex]formula units.

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