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iren2701 [21]
3 years ago
15

How many moles of atoms are in 45.0g of gold

Chemistry
2 answers:
Inessa05 [86]3 years ago
6 0
Hello!

45.0g \: au \times \frac{1 \: mol \: au}{196.97 \: g \: au} \: = 0.228 \: mol \: au

And there's your answer! ^ I really hope this helped you out! :)
DiKsa [7]3 years ago
3 0
Hi how are nice to meet you
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For iii:22.3mL
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The symbol for gold is<br> a. go.<br> b. g.<br> c. au.<br> d. au.
Zolol [24]
Au is the symbol for gold. It comes from Latin word for gold, which is Aurunum. Its atomic number is 79. 
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In class we derived the Gibbs energy of mixing for a binary mixture of perfect gases. We also discussed that the same result is
GaryK [48]

Answer:

Attached below

Explanation:

Free energy of mixing = ΔGmix = Gf - Gi

attached below is the required derivation of the

<u>a) Molar Gibbs energy of mixing</u>

ΔGmix = Gf - Gi

hence : ΔGmix = ∩RT ( X1 In X1 + X2 In X2 + X3 In X3 + ------- )

<u>b) molar excess Gibbs energy of mixing</u>

Ni = chemical potential of gas

fi = Fugacity

N°i = Chemical potential of gas when Fugacity = 1

ΔG = RT In ( a2 / a1 )  

4 0
3 years ago
A 400 ml sample of gas is heated from -20 c to 60
Viktor [21]
Using p1v1/t1=p2v2/t2
p1=50
p2=225
v1=400ml
v2=?
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7.9=0.7xv2
v2=7.9/0.7
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6 0
3 years ago
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The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10
Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

                   2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0        125 kPa

At t = teq     125 - 2x      4x        x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 2.8\times 10^{-3} min⁻¹

Initial concentration [A_0] = 125 kPa

Final concentration [A_t] = 91 kPa

Time = ?

Applying in the above equation, we get that:-

91=125e^{-2.8\times 10^{-3}\times t}

125e^{-2.8\times \:10^{-3}t}=91

-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)

t=113\ min

3 0
2 years ago
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