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3 years ago

At high temperatures, steam decomposes into its elements according to the following equation.

2 answers:

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_Award brainliest if helped!
20% steam converted, 0.8atm of steam remains
Converted to 0.2atm H2(g) and 0.1atm O2(g)
Partial Pressure of
H2O(g) =0.8/1.1 =0.727atm
H2(g) = 0.2/1.1 =0.181atm
O2(g) = 0.1/1.1 = 0.091atm

lina2011 [118]3 years ago

5 0

Answer:

Explanation:

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3 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

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Answer: The number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

Explanation:

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

$N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}$

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$N_{Au}$ = number of gold atoms per cubic centimeters

$N_A$ = Avogadro's number = $6.022\times 10^{23}atoms/mol$

$C_{Au}$ = Mass percent of gold in the alloy = 42 %

$\rho_{Au}$ = Density of pure gold = $19.32g/cm^3$

$\rho_{Ag}$ = Density of pure silver = $10.49g/cm^3$

$M_{Au$ = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

$N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3$

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3 years ago

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