Which option contains a $2 \cdot g$ $2g$ mass of sodium atoms.... $1 .$ $1.$ $5.24 \times 10^{22} \cdot sodium atoms?$ $5.24xx10^22"sodium atoms?"$ $2 .$ $2.$ $6.022 \times 10^{24} \cdot sodium atoms?$ $6.022xx10^24"sodium atoms?"$ $3 .$ $3.$ $3.12 \times 10^{23} \cdot sodium atoms?$ $3.12xx10^23"sodium atoms?"$ $4 .$ $4.$ $5.24 \times 10^{22} \cdot sodium atoms?$ $5.24xx10^22*"sodium atoms?"$

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Which option contains a $2 \cdot g$ $2g$ mass of sodium atoms.... $1 .$ $1.$ $5.24 \times 10^{22} \cdot sodium atoms?$ $5.24xx10^22"sodium atoms?"$ $2 .$ $2.$ $6.022 \times 10^{24} \cdot sodium atoms?$ $6.022xx10^24"sodium atoms?"$ $3 .$ $3.$ $3.12 \times 10^{23} \cdot sodium atoms?$ $3.12xx10^23"sodium atoms?"$ $4 .$ $4.$ $5.24 \times 10^{22} \cdot sodium atoms?$ $5.24xx10^22*"sodium atoms?"$

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Answer 1 · 2017-06-08T19:41:14

I would say $option 1........$ $"option 1........"$

Explanation:

We know that $1 \cdot m o l$ $1*mol$ contains $6.022 \times 10^{23}$ $6.022xx10^23$ individual sodium atoms, and has a mass of $22.99 \cdot g$ $22.99*g$ .

And thus for $1 .$ $1.$ there is a $2 \cdot g$ $2*g$ mass of $N a$ $Na$ .

For $2 .$ $2.$ there is a $22.99 \cdot g$ $22.99*g$ mass of $N a$ $Na$ .

For $3 .$ $3.$ there is a ..............................

$\frac{3 \times 10^{23} \cdot sodium atoms}{6.022 \times 10^{23} \cdot sodium atoms \cdot m o l^{- 1}} \times 22.99 \cdot g \cdot m o l^{- 1}$ $(3xx10^23*"sodium atoms")/(6.022xx10^23*"sodium atoms"*mol^-1)xx22.99*g*mol^-1$ mass of $N a$ $Na$ , approx. $11.5 \cdot g$ $11.5*g$ .

For $4 .$ $4.$ there is a $8 \cdot g$ $8*g$ mass of sodium............

And so, clearly,..................Capisce?

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