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3 years ago

8

Calculate the frequency required to eject electrons from gold with a work function of 8.76 x 10-19 J

1 answer:

Savatey [412]3 years ago

5 0

$hf=W+\frac{mv^{2}}{2}\\\\
f=\frac{W+\frac{mv^{2}}{2}}{h}\\\\\

W=8,76*10^{-19}J\\
m=9,11*10^{-31}kg\\
v=3*10^{8}\frac{m}{s}\\
h=6,63*10^{-34}Js\\\\\\
f=\frac{8,76*10^{-19}J+\frac{9,11*10^{-31}kg*(3*10^{8}\frac{m}{s})^{2}}{2}}{6,63*10^{-34}Js}=7,5*10^{19}Hz$

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

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Answer : The mass of $MnCO_3$ required are, 35 kg

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First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

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Let the mass of $MnCO_3$ be, 'x' grams.

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As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

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The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

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