a. 44 b. 2.64times 10^25g C. 7.31g d. 64g 86. How many molecules of SO_(2) in 1.60 moles? a. 102.1times 10^1 b. 9.63times 10^23 C. 3.76times 10^23 d. 7.62times 10^1 87. What is the percent composition of Cr in chromium oxide (Cr_(2)O_(3)) a. 31.6% b. 68.4% C. 34.2% d. 76.4% 88. Sodium nitrite (NaNO_(2)) is commonly used in processed meat as a preservative. It preserves the red pink color in meats and delays the development of botulism. Find the percent composition of sodiur sodium nitrite. a. 20.30% b. 46.38% C. 33.32% d. 69.00%

86. To find the number of molecules of $SO_{2}$ in 1.60 moles, we can use Avogadro's number, which is $6.022 \times 10^{23}$ molecules per mole.<br /><br />Number of molecules = (Number of moles) x (Avogadro's number)<br />Number of molecules = 1.60 moles x $6.022 \times 10^{23}$ molecules/mole<br />Number of molecules = $9.63 \times 10^{23}$ molecules<br /><br />Therefore, the correct answer is b. $9.63\times 10^{23}$.<br /><br />87. To find the percent composition of Cr in chromium oxide $(Cr_{2}O_{3})$, we need to calculate the molar mass of the compound and then determine the mass of Cr in one mole of the compound.<br /><br />Molar mass of $Cr_{2}O_{3}$ = (2 x molar mass of Cr) + (3 x molar mass of O)<br />Molar mass of $Cr_{2}O_{3}$ = (2 x 52.00 g/mol) + (3 x 16.00 g/mol)<br />Molar mass of $Cr_{2}O_{3}$ = 152.00 g/mol<br /><br />Percent composition of Cr = (Mass of Cr / Molar mass of $Cr_{2}O_{3}$) x 100%<br />Percent composition of Cr = (2 x 52.00 g/mol) / 152.00 g/mol x 100%<br />Percent composition of Cr = 68.4%<br /><br />Therefore, the correct answer is b. $68.4\%$.<br /><br />88. To find the percent composition of sodium in sodium nitrite $(NaNO_{2})$, we need to calculate the molar mass of the compound and then determine the mass of Na in one mole of the compound.<br /><br />Molar mass of $NaNO_{2}$ = (1 x molar mass of Na) + (1 x molar mass of N) + (2 x molar mass of O)<br />Molar mass of $NaNO_{2}$ = (1 x 22.99 g/mol) + (1 x 14.01 g/mol) + (2 x 16.00 g/mol)<br />Molar mass of $NaNO_{2}$ = 69.00 g/mol<br /><br />Percent composition of Na = (Mass of Na / Molar mass of $NaNO_{2}$) x 100%<br />Percent composition of Na = 22.99 g/mol / 69.00 g/mol x 100%<br />Percent composition of Na = 33.32%<br /><br />Therefore, the correct answer is c. $33.32\%$.