Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
A 0.750 g sample of tin is oxidized with nitric acid to form tin oxide. if the original tin sample gained 0.21 g of oxygen, the empirical formula of tin oxide is SnO₂.
A 0.750 g sample of tin is oxidized with nitric acid to form tin oxide. if the original tin sample gained 0.21 g of oxygen, the mass of tin oxide is:
[tex]mSn_xO_y = mSn + mO = 0.750 g + 0.21 g = 0.96 g[/tex]
Then, we will calculate the mass percent of each element in the oxide.
[tex]\% Sn = \frac{0.750g}{0.96g} \times 100\% = 78\%\\\\\% O = \frac{0.21g}{0.96g} \times 100\% = 22\%[/tex]
Next, we will divide each percentage by the atomic mass of the element.
[tex]Sn: 78/118.71 = 0.66\\\\O: 22/16.00 = 1.4[/tex]
Finally, we will divide both numbers by the smallest one, i.e. 0.66.
[tex]Sn: 0.66/0.66 = 1\\\\O: 1.4/0.66 \approx 2[/tex]
The empirical formula of tin oxide is SnO₂.
A 0.750 g sample of tin is oxidized with nitric acid to form tin oxide. if the original tin sample gained 0.21 g of oxygen, the empirical formula of tin oxide is SnO₂.
Learn more: https://brainly.com/question/1363167

Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.