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3 years ago

How many moles of atoms are in 45.0g of gold

2 answers:

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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PLEASE HELP BEFORE 7 A.M. PACIFIC TIME SEE ATTACHED.

anastassius [24]

For i: 33mL
For ii: 87-88mL
For iii:22.3mL

5 0

2 years ago

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The symbol for gold is a. go. b. g. c. au. 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.

7 0

2 years ago

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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

a) Molar Gibbs energy of mixing

ΔGmix = Gf - Gi

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

b) molar excess Gibbs energy of mixing

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=?
t1=-20=253k
t2=60=333k
50x400/253=225xv2/333
7.9=0.7xv2
v2=7.9/0.7
v2=11.3ml

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