What mass of silver (Ag) contains the same number of atoms as 50.6 g of boron (B)?

Answer 1

Approx. #500*g#.............

#"Moles of boron"=(50.6*g)/(10.81*g*mol^-1)=4.68*mol#
Now remember that the mole is simply a number, i.e. #6.022xx10^23*mol^-1#... And to get the same number of silver atoms, all I have to do is multiply this molar quantity by the molar mass of silver, #107.9*g*mol^-1#.....
And so, #107.9*g*mol^-1xx4.68*mol-=??#
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Answer 2

To find the mass of silver containing the same number of atoms as 50.6 g of boron, you would first calculate the molar mass of boron, then use Avogadro's number to find the number of atoms, and finally, divide that number by Avogadro's number and multiply by the molar mass of silver. The molar mass of boron is approximately 10.81 g/mol, and Avogadro's number is (6.022 \times 10^{23}) atoms/mol. Therefore, the mass of silver containing the same number of atoms as 50.6 g of boron would be approximately 107.87 g.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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