If the mass of a single electron is $9.10938 \times 10^{- 31} \cdot k g$ $9.10938xx10^-31kg$ , how many electrons constitute a mass of $1 \cdot g$ $1g$ ?

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If the mass of a single electron is $9.10938 \times 10^{- 31} \cdot k g$ $9.10938xx10^-31kg$ , how many electrons constitute a mass of $1 \cdot g$ $1g$ ?

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Answer 1 · 2017-09-16T12:13:36

You could have quoted the $electronic mass$ $"electronic mass"$ , ..........

Explanation:

And the mass of electron is .................. $9.10938 \times 10^{- 31} \cdot k g \equiv 9.10938 \times 10^{- 28} \cdot g$ $9.10938xx10^-31*kg-=9.10938xx10^-28*g$

And thus number of electrons is.....

$\frac{1.000 \cdot g}{9.10938 \times 10^{- 28} \cdot g} = 1.098 \times 10^{27}$ $(1.000*g)/(9.10938xx10^-28*g)=1.098xx10^27$ $individual electrons.$ $"individual electrons."$

And note that the charge of this quantity of electrons, i.e. $−1.602×10^(-19) C*"electron"^-1xx1.098xx10^27*"electron"$ would be phenomenonally GREAT.

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If the mass of a single electron is $9.10938 \times 10^{- 31} \cdot k g$ $9.10938xx10^-31kg$ , how many electrons constitute a mass of $1 \cdot g$ $1g$ ?

1 Answer

Explanation:

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If the mass of a single electron is 9.10938xx10^-31*kg9.10938×10−31⋅kg9.10938xx10^-31*kg, how many electrons constitute a mass of 1*g1⋅g1*g?

1 Answer

Explanation:

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If the mass of a single electron is $9.10938 \times 10^{- 31} \cdot k g$ $9.10938xx10^-31kg$ , how many electrons constitute a mass of $1 \cdot g$ $1g$ ?