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AlladinOne [14]
3 years ago
11

A student combines 364.6 g of HCl with 80 g of NaOH in 5 L of water. What additional volume of H2O must be added to this mixture

to yield a solution with a pH of 1? Note that the molar mass of HCl is 35.46 g/mole, while that of NaOH is 40 g/mole.
Chemistry
1 answer:
Jobisdone [24]3 years ago
6 0

Answer:

75L of additional water to have a pH 1 solution

Explanation:

The reaction of HCl With NaOH is:

HCl + NaOH → H₂O + NaCl

By using molar mass of each reactant you can know how many moles will react, thus:

HCl: 364.6g HCl ₓ (1mol / 36.46g) = 10 moles HCl

NaOH: 80g NaOH ₓ (1mol / 40g) = 2 moles NaOH

That means after the reaction will remain in solution, 10-2 = 8 moles of HCl = 8 moles of H⁺ (In water, HCl dissociates as H⁺ and Cl⁻ ions).

A solution with pH = 1 contains:

pH = -log [H⁺]

1 = -log [H⁺]

0.1M = [H⁺]

As molarity, M is the ratio between moles and liters and you want a solution 0.1M having 8 moles of H⁺ you require:

0.1M = 8 moles H⁺ / 80L

As the student combines the solution with 5L of water, you require

<h3>75L of additional water to have a pH 1 solution</h3>
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Fe2O3(s) + 3CO(g) ---&gt; 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast
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Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles

\text{Moles of} CO=\frac{260}{28}=9.3moles

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)

According to stoichiometry :

1 mole of Fe_2O_3 require 3 moles of CO

Thus 2.8 moles of Fe_2O_3 will require=\frac{3}{1}\times 2.8=8.4moles  of CO

Thus Fe_2O_3 is the limiting reagent as it limits the formation of product and CO is the excess reagent.

As 1 mole of Fe_2O_3 give = 2 moles of Fe

Thus 2.8 moles of Fe_2O_3 give =\frac{2}{1}\times 2.8=5.6moles of Fe

Mass of Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%

Therefore, the percent yield is, 91.8%

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<span>Gold have a single outer electron. This seems disadvantageous, energy-wise, until you look at the orbitals the electrons are in. The lone electron is in an S-orbital. This orbital is thus half full (since s-orbitals can contain 2 electrons), whereas all the other inner orbitals in silver and gold are filled, and hence exceptionally stable. After a full orbital, the next most stable orbital is a half full one. </span>
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1. An electric power plant uses energy from burning coal to generate steam at 450 °C. The plant is cooled by 20 °C water
kap26 [50]

Answer:

Explanation:

Efficiency of the electric power plant is e=1-\frac{T_{2}}{T_{1}}

Here Temperature of hot source T_{1} = 450^{o}C=450+273=723 K

and Temperature of sink T_{1} = 20^{o}C=20+273=293 K

Hence the efficiency is  e=1-\frac{293}{723}=0.5947=59.47%

Now another formula for thermal efficiency Is

e_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}

Here QI is the of heat taken from source 100 MJ ; Q2 of heat transferred to the sink (river) to be found

W is the of work done and W = QI -Q2

Hence Frome_{therm}=\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{W}{Q_{1}}

W=e(Q_{1})=(0.5947)(100)=59.47MJ

Hence the of heat transferred to the river Is Q_{2} -W = (100 -59.47=40.53

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