goles Test 81. The GFM of magnesium chloride, MgCl_(2) is: a. 5908g/mol b. 95.2g/mol C 125.8g/mol d. 76.4g/mol 82. What is the percent composition of aluminum in aluminum sulfat a. 54.0% b. 28.1% C. 15.8% d. 84.2% 83. How many grams are in 2.00 moles of sodium sulfate (Na_(2)SO_(4) a. 2849 b. 2389 C. 0.0141 g d. 0.0170 g 84. Which of the following is an empirical formula? a. CH_(3) b. C_(6)H_(12)O_(6) N_(2)O_(4) d. C_(5)H_(10) 85. How many grams are in 6.02times 10^23 molecules of CO_(2) a. 449 b. 2.64times 10^25g C. 7.319 d. 649 60 moles? Name

81. The molar mass of magnesium chloride, $MgCl_{2}$, is calculated by adding the molar masses of magnesium (Mg) and chlorine (Cl). The molar mass of Mg is 24.305 g/mol, and the molar mass of Cl is 35.453 g/mol. Since there are two chlorine atoms in $MgCl_{2}$, the molar mass is:<br /><br />\[ \text{Molar mass of } MgCl_{2} = 24.305 \, \text{g/mol} + 2 \times 35.453 \, \text{g/mol} = 24.305 \, \text{g/mol} + 70 \text{g/mol} = 95.211 \, \text{g/mol} \]<br /><br />Therefore, the correct answer is:<br />b. $95.2g/mol$<br /><br />82. The percent composition of aluminum in aluminum sulfate, $Al_{2}(SO_{4})_{3}$, is calculated by dividing the total mass of aluminum by the molar mass of aluminum sulfate and then multiplying by 100. The molar mass of aluminum is 26.98 g/mol, and the molar mass of aluminum sulfate is:<br /><br />\[ \text{Molar mass of } Al_{2}(SO_{4})_{3} = 2 \times 26.98 \, \text{g/mol} + 3 \times (32.06 \, \text{g/mol} + \times 16.00 \, \text{}) = 53.96 \, \text{g/mol} + 3 \times 96.06 \, \text{g/mol} = 53.96 \, \text{g/mol} + 288.18 \, \text{g/mol} = 342.14 \, \text{g/mol} \]<br /><br />The percent composition of aluminum is:<br /><br />\[ \text{Percent composition of Al} = \left( \frac{2 \times 26.98 \, \text{g/mol}}{342.14 \, \text{g/mol}} \right) \times 100 = \left( \frac{53.96 \, \text{g/mol}}{342.14 \, \text{g/mol}} \right) 100 \approx 15.8\% \]<br /><br />Therefore, the correct answer is:<br />c. $15.8\%$<br /><br />83. The molar mass of sodium sulfate, $Na_{2}SO_{4}$, is calculated by adding the molar masses of sodium (Na), sulfur (S), and oxygen (O). The molar mass of Na is 22.99 g/mol, the molar mass of S is 32.06 g/mol, and the molar mass of O is 16.00 g/mol. Since there are two sodium atoms, one sulfur atom, and four oxygen atoms in $Na_{2}SO_{4}$, the molar mass is:<br /><br />\[ \text{Molar mass of } Na_{2}SO_{4} = 2 \times 22, \text{g/mol} + 32.06 \, \text{g/mol} + 4 \times 16.00 \, \text{g/mol} = 45.98 \, \text{g/mol} + 32.06 \, \text{g/mol} + 64.00 \, \text{g/mol} = 142.04 \, \text{g/mol} \]<br /><br />To find the mass of 2.00 moles of $Na_{2}SO_{4}$, we multiply the molar mass by the number of moles:<br /><br />\[ \text{Mass} = 2.00 \, \text{moles} \times 142.04 \, \text{g/mol} = 284.08 \, \text{g} \]<br /><br />Therefore, the correct answer is2849<br /><br />84. An empirical formula represents the simplest whole-number ratio of atoms in a compound. Among the given options, $CH_{3}$ is the simplest whole-number ratio of carbon (C) and hydrogen (H), making it an empirical formula.<br /><br />Therefore, the correct answer is:<br />a. $CH_{3}$<br /><br />85. To find the mass of $6.02 \times 10^{23}$ molecules of $CO_{2}$, we first need to convert the number of molecules to moles using Avogadro's number, which is $6.02 \times 10^{23}$ molecules/mol. Since the molar mass of $CO_{2}$ is 44.01 g/mol, the mass of $6.02 \times 10^{23}$ molecules of $CO_{2}$ is:<br /><br />\[ \text{Mass