Question #70cfb

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Question #70cfb

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Answer 1 · 2017-05-16T03:03:20

$10^{23} p r o t o n s$ $10^(23) prot ons$ (order of magnitude, you asked)

Explanation:

If one proton has a mass of $1.67 \times 10^{- 27} k g$ $1.67 xx 10^-27 kg$ , and there are $10^{3}$ $10^3$ grams in one kilogram, one gram of protons contains

$1 g (\frac{1 k g}{10^{3} g}) (\frac{1 p r o t o n}{1.67 \times 10^{- 27} k g}) = 6.00 \times 10^{23} p r o t o n s$ $1 g((1kg)/(10^3g))((1 prot on)/(1.67 xx 10^(-27)kg)) = 6.00 xx 10^23 prot ons$

and the order of magnitude is thus in $10^{23}$ $10^23$

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Question #70cfb

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