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Problem 36

Calculate the following quantities: $$ \begin{a…

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Problem 35

Calculate the following quantities:
$$
\begin{array}{l}{\text { (a) mass, in grams, of } 0.105 \text { mol sucrose }\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)} \\ {\text { (b) moles of } \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} \text { in } 143.50 \mathrm{g} \text { of this substance }} \\ {\text { (c) number of molecules in } 1.0 \times 10^{-6} \mathrm{mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}} \\ \text { (d) number of }
\mathrm{N} \text { atoms in } 0.410 \mathrm{mol} \mathrm{NH}_{3}\end{array}
$$

Answer

a) 35.94 $\mathrm{g}$
b) 0.76 $\mathrm{mol}$
c) See explanation
d) 2.47$\cdot 10^{23}$ atoms $N$


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Video Transcript

So for this cushion, we need the conversion factor for one mole off seat. Well, it's 20 to 11 which is equal do 300 42.2 grounds converting it into Mul. The mosque, or zero point 105 mole off Sue Cruz is 35.9 ground. Using the same formula as above, we console all off the options. The mosque for 0.757 moles off zinc and no tree too is 143.50 grams violet 6.0 into 10 days to pass 17 molecules off. See its trees here too, which are present in 1.0 into 10 days to part minus six more, though CS three CS to what for the last one 2.47 into 10 days to put 23 atoms off. My true Jenna present in 0.410 moons off the next three for last option. One molecule off in its tree is equal to one Adam off nitwit

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