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3 years ago

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An economical substitute for vitreous silica is a high-silica glass made by leaching the B2O3-rich phase from a two-phase borosi

licate glass. (The resulting porous microstructure is densified by heating.) A typical starting composition is 81 wt % SiO2, 4 wt % Na2O, 2 wt % Al2O3, and 13 wt % B2O3. A typical final composition is 96 wt % SiO2, 1 wt % Al2O3, and 3 wt % B2O3. How much product (in kilograms) would be produced from 50 kg of starting material, assuming no SiO2 is lost by leaching

1 answer:

stich3 [128]3 years ago

7 0

Answer:48kg of SiO2, 0.5kg of Al2O3, and 1.5kg of B2O3

Will be the final product

Explanation:

I) 96wt% of SiO2 will amount to 96/100*50 = 0.96*50=48kg of SiO2

ii) 1wt% of Al2O3 will amount to 1/100*50 = 0.01*50=0.5kg of Al2O3

III) 3wt% of B2O3 will amount to 3/100*50 = 0.03*50=1.5kg of B2O3..

The overall product form 48+ 0.5+1.5= 50kg

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Americans combined drive about 4.0 x 109 kilometers a day and get an average of 20 miles per gallon of gasoline. For each kilogr

Nimfa-mama [501]

Answer:

$303,882.84649 kg\times 3=9.12\times 10^5 kg$ of carbon dioxide gas.

Explanation:

Average distance covered by Americans in a day= $4.0\times 10^9 km$

1 day = 24 × 60 min = 1,440 min

Average distance covered by Americans in a minute= $\frac{4.0\times 10^9 km}{1,440}=2,777,777.78 km$

Average mileage of the car = 20 miles/gal = 32.18 km/gal

1 mile = 1.609 km

20 miles = 20 × 1.609 km = 32.18 km

Volume of gasoline used in minute = $\frac{2,777,777.78 km}{32.18 km/gal}$

$V=86,320.00 gal$

$V=86,320.00\times 3.7854 L$

(1 L = 1000 mL)

$V=86,320.00\times 3.7854 \times 1000 mL=326,755,748.91 mL$

Mass of 86,320.00 gallons of gasoline = m

Density of the gasoline = d = $0.93 g/cm^3=0.93 g/mL$

$1 mL= 1 cm^3$

$m=d\times V=0.93 g/mL\times 326,755,748.91 mL$

$m=303,882,846.49 g=303,882.84649 kg$

1 kilogram of gasoline gives 3 kg of carbon dioxde gas .

Then 303,882.84649 kg of gasoline will give :

$303,882.84649 kg\times 3=9.12\times 10^5 kg$ of carbon dioxide gas.

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3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

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

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

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If the concentrated form of blood expander is 9.0 mol/L and one of your companions, a nurse, tells you that you need to have a f

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Answer:

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Explanation:

Use the dilution equation M1V1 = M2V2

M1 = 9.0 M

V1 = This is what we're looking for.

M2 = 0.145 M

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Solve for V1 --> V1 = M2V2/M1

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AleksAgata [21]

Explanation:

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3 years ago

How many atoms of Fe are in 45.00 grams of Fe?

Licemer1 [7]

Answer:

4.852*10^23

Explanation:

Use dimensional analysis.

First turn grams to moles using the molar mass of Fe (you can't convert to atoms from grams).

Then turn moles into atoms using Avogadro's constant.

45g Fe/55.85 g/mol Fe*6.022*10^23 atoms Fe

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3 years ago