Given :
Number of molecules of hydrogen peroxide, N = 4.5 × 10²².
To Find :
The mass of given molecules of hydrogen peroxide.
Solution :
We know, 1 mole of every compound contains Nₐ = 6.022 × 10²³ molecules.
So, number of moles of hydrogen peroxide is :

Now, mass of hydrogen peroxide is given as :
m = n × M.M
m = 0.0747 × 34 grams
m = 2.54 grams
Hence, this is the required solution.
Answer:
A. 266g/mol
Explanation:
A colligative property of matter is freezing point depression. The formula is:
ΔT = i×Kf×m <em>(1)</em>
Where:
ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g
Replacing in (1):
0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg
<em>0,0614 = mol of solute</em>.
As molar mass is defined as grams per mole of substance and the compound weights 16,0g:
16,0g / 0,0614 mol = 261 g/mol ≈ <em>A. 266g/mol</em>
I hope it helps!
Answer:

Explanation:
Given that,
The wavelength of a microwave is 7.42 mm or 0.00742 m
No. of photons, n = 359
We need to find the energy produced by this no of photons. It can be given by the formula as follows :

or

So, the required energy is
.
Answer:

Explanation:
Hello!
In this case, since the definition of entropy in a random mixture is:
![\Delta S=-n_TR\Sigma[x_i*ln(x_i)]](https://tex.z-dn.net/?f=%5CDelta%20S%3D-n_TR%5CSigma%5Bx_i%2Aln%28x_i%29%5D)
For this silver-gold mixture we write:
![\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]](https://tex.z-dn.net/?f=%5CDelta%20S%3D-%28n_%7BAu%7D%2Bn_%7BAg%7D%29R%5CSigma%5B%5Cfrac%7Bn_%7BAu%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%2Aln%28%5Cfrac%7Bn_%7BAu%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%29%2B%5Cfrac%7Bn_%7BAg%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%2Aln%28%5Cfrac%7Bn_%7BAg%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%29%5D)
By knowing the moles of gold:

It is possible to write the aforementioned formula in terms of the variable
representing the moles of silver:
![20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]](https://tex.z-dn.net/?f=20%5Cfrac%7BJ%7D%7Bmol%7D%3D-%280.508%2Bx%298.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%5CSigma%5B%5Cfrac%7B0.508%7D%7B0.508%2Bx%7D%20%2Aln%28%5Cfrac%7B0.508%7D%7B0.508%2Bx%7D%20%29%2B%5Cfrac%7Bx%7D%7B0.508%2Bx%7D%20%2Aln%28%5Cfrac%7Bx%7D%7B0.508%2Bx%7D%20%29%5D)
Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

So the mass is:

Best regards!