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How many atoms of C are in 0.185 mol of (C3H5)2O?
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is this \[\large (C_3 H_5)_2 O?\]
Yes
\[0.185 \; \cancel{\text{mol} \;(C_3 H_5)_2 O} \times \frac{6 \; \cancel{\text{mol} \; C}}{1 \; \cancel{\text{mol} \;(C_3 H_5)_2 O}} \times \frac{6.022 \times 10^{23} \; \text{atoms} \; C}{1 \; \cancel{\text{mol} \; C}}\] do you get that?
Yes, thank you. I came up with 6.68 X 10^23 mol C.
i have no way of verifying that sorry. i have no calculator in hand
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