What are the molecular weights of the following compounds? (all masses must be to nearest hundredth) 1) NaOH 2) H_(3)PO_(4) 3) H_(2)O 4) Mn_(2)Se_(7) 5) MgCl_(2) 6) (NH_(4))_(2)SO_(4)

1) The molecular weight of NaOH is calculated by adding the atomic weights of Sodium (Na), Oxygen (O), and Hydrogen (H). Sodium has an atomic weight of 22.99, Oxygen has an atomic weight of 16.00, and Hydrogen has an atomic weight of 1.01. Therefore, the molecular weight of NaOH is 22.99 + 16.00 + 1.01 = 40.00.<br />2) The molecular weight of $H_{3}PO_{4}$ is calculated by adding the atomic weights of Hydrogen (H), Phosphorus (P), and Oxygen (O). Hydrogen has an atomic weight of 1.01, Phosphorus has an atomic weight of 30.97, and Oxygen has an atomic weight of 16.00. Therefore, the molecular weight of $H_{3}PO_{4}$ is 3(1.01) + 30.97 + 4(16.00) = 97.00.<br />3) The molecular weight of $H_{2}O$ is calculated by adding the atomic weights of Hydrogen (H) and Oxygen (O). Hydrogen has an atomic weight of 1.01, and Oxygen has an atomic weight of 16.00. Therefore, the molecular weight of $H_{2}O$ is 2(1.01) + 16.00 = 18.02.<br />4) The molecular weight of $Mn_{2}Se_{7}$ is calculated by adding the atomic weights of Manganese (Mn) and Selenium (Se). Manganese has an atomic weight of 54.94, and Selenium has an atomic weight of 78.96. Therefore, the molecular weight of $Mn_{2}Se_{7}$ is 2(54.94) + 7(78.96) = 776.64.<br />5) The molecular weight of $MgCl_{2}$ is calculated by adding the atomic weights of Magnesium (Mg) and Chlorine (Cl). Magnesium has an atomic weight of 24.31, and Chlorine has an atomic weight of 35.45. Therefore, the molecular weight of $MgCl_{2}$ is 24.31 + 2(35.45) = 95.21.<br />6) The molecular weight of $(NH_{4})_{2}SO_{4}$ is calculated by adding the atomic weights of Nitrogen (N), Hydrogen (H), Sulfur (S), and Oxygen (O). Nitrogen has an atomic weight of 14.01, Hydrogen has an atomic weight of 1.01ulf an atomic weight of 32.07, and Oxygen has an atomic weight of 16.00. Therefore, the molecular weight of $(NH_{4})_{2}SO_{4}$ is 2(14.01 + 4(1.01)) + 32.07 + 4(16.00) = 132.14.