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likoan [24]
9 months ago
5

5 g of gold and 25 g of silver are mixed to form a single-phase ideal solid solution. (a) how many moles of solution are there?

(b) what are the mole fractions of gold and silver?
Chemistry
1 answer:
worty [1.4K]9 months ago
3 0

Mole fraction for silver= 0.2317643/0.3079193

                                     = 0.7526787

Mole fraction for gold=0.076155/0.3079193

                                    =0.2473212

What is Molar Fraction?

The mole fraction is the product of the number of molecules of a specific component in a mixture and its total molecular weight. It serves as a means of indicating how concentrated a solution is.

It is a unit of concentration, defined as the product of the moles of a component and the moles of the entire solution. When all the parts of a solution are summed up, their mole fraction equals 1.

molar mass of silver = 107.8682g/mol

Molar mass of gold= 196.96657g/mol

Therefore mole = mass/molar mass

For silver: 25g/107.8682g/mol = 0.2317643mol

For gold: 15g/196.96657g/mol= 0.076155mol

Total number of mole= 0.2317643+0.076155

                                    = 0.30791193mol

Mole fraction for silver= 0.2317643/0.3079193

                                      = 0.7526787

Mole fraction for gold=0.076155/0.3079193

                                     =0.2473212

Learn more about Mole Fraction from given link

brainly.com/question/28590918

#SPJ4

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Step 1:
           
Write molecular Formula,

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Assign Oxidation number to Na (which is +1) and O (which is -2), and X (unknown) for P.
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                                                X ÷ 3  =  -5 + 20
                                
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The initial activity of 37Ar is 8540 disintegrations per minute. After 10.0 days, the activity is 6990 disintegrations per minut
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Approximately 3318 disintegrations per minute.

Explanation:

The activity A of a radioactive decay at time t can be found with the following equation:

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In this equation,

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  • A_0 is the initial activity of the decay. For this question, A_0 = \rm 8540\; min^{-1}.
  • The decay constant \lambda of this sample needs to be found.

The decay constant here can be found using the activity after 10 days. As long as both times are in the same unit (days in this case,) conversion will not be necessary.

A(10) = A_0\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}= (\mathrm{8540\; min^{-1}})\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}.

A(10) = \rm 6690\; min^{-1}.

\displaystyle \frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}} = \mathrm{e}^{\rm -\lambda \times 10\;day}

Apply the natural logarithm to both sides of this equation.

\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)} = \ln{\left(\mathrm{e}^{\rm -\lambda \times 10\;day}\right)}.

\displaystyle \rm -\lambda \cdot (10\;day) = \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}.

\displaystyle \rm \lambda= \rm \frac{\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}}{-10 \; day} \approx 0.0200280\; day^{-1}.

Note that the unit of the decay constant \lambda is \rm day^{-1} (the reciprocal of days.) The exponent -\lambda \cdot t should be dimensionless. In other words, the unit of t should also be days. This observation confirms that there's no need for unit conversion as long as the two times are in the same unit.

Apply the equation for decay activity at time t to find the decay activity after 47.2 days.

\displaystyle \begin{aligned}A(t)& = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}\\&\approx \rm \left(8540\; min^{-1}\right)\cdot \mathrm{e}^{-0.0200280\; day^{-1}\times 47.2\;day}\\&\approx \rm 3318\; min^{-1}\end{aligned}.

By dimensional analysis, the unit of activity here should also be disintegrations per minute. The activity after 47.2 days will be approximately 3318 disintegrations per minute.

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