To find the mass of 21.3 moles of lithium (Li), you can use the molar mass of lithium, which is approximately 6.94 g/mol. Multiply the number of moles by the molar mass: (21.3 , \text{mol} \times 6.94 , \text{g/mol} = 147.58 , \text{g}). Therefore, the mass of 21.3 moles of lithium is approximately 147.58 grams.
The atomic mass of Li, which stands for lithium, is approximately 6.94 atomic mass units (u).
The chemical formula for lithium oxide is ( \text{Li}2\text{O} ), not ( \text{Li}{20} ). To calculate the molar mass of lithium oxide, you add the molar masses of its components: lithium (Li) has a molar mass of about 6.94 g/mol, and oxygen (O) has a molar mass of about 16.00 g/mol. Thus, the molar mass of ( \text{Li}_2\text{O} ) is ( 2 \times 6.94 , \text{g/mol} + 16.00 , \text{g/mol} = 29.88 , \text{g/mol} ).
The element with the least mass in Period 2 is Lithium (Li). It has an atomic mass of approximately 6.94 g/mol
To find the number of moles in each elemental sample, we use the formula: moles = mass (g) / molar mass (g/mol). Argon (Ar): ( \text{molar mass} = 39.95 , \text{g/mol} ), so ( \text{moles} = 11.8 , \text{g} / 39.95 , \text{g/mol} \approx 0.295 , \text{mol} ). Zinc (Zn): ( \text{molar mass} = 65.38 , \text{g/mol} ), so ( \text{moles} = 3.55 , \text{g} / 65.38 , \text{g/mol} \approx 0.0543 , \text{mol} ). Tantalum (Ta): ( \text{molar mass} = 180.95 , \text{g/mol} ), so ( \text{moles} = 26.1 , \text{g} / 180.95 , \text{g/mol} \approx 0.144 , \text{mol} ). Lithium (Li): ( \text{molar mass} = 6.94 , \text{g/mol} ), so ( \text{moles} = 0.211 , \text{g} / 6.94 , \text{g/mol} \approx 0.0304 , \text{mol} ).
To find the mass of 0.50 mol of aluminum foil, we first need the molar mass of aluminum, which is approximately 27 g/mol. Therefore, the mass can be calculated using the formula: mass = moles × molar mass. For 0.50 mol of aluminum, the mass would be 0.50 mol × 27 g/mol = 13.5 grams.
The molar mass of Li = 6.941 g/mol
The atomic mass of Li, which stands for lithium, is approximately 6.94 atomic mass units (u).
The chemical formula for lithium oxide is ( \text{Li}2\text{O} ), not ( \text{Li}{20} ). To calculate the molar mass of lithium oxide, you add the molar masses of its components: lithium (Li) has a molar mass of about 6.94 g/mol, and oxygen (O) has a molar mass of about 16.00 g/mol. Thus, the molar mass of ( \text{Li}_2\text{O} ) is ( 2 \times 6.94 , \text{g/mol} + 16.00 , \text{g/mol} = 29.88 , \text{g/mol} ).
Lithium has a molar mass of 6.94 g/mol. Oxygen has a molar mass of 16.00 g/mol. Since Lithium Oxide has 2 Lithium atoms, the molar mass is: (6.94 x 2) + 16.00 = 29.88 g/mol.
How many lithium atoms are in 10.56 g of lithium
The molar mass of LiOH is 23.95 g/mol for Li, 15.999 g/mol for O, and 1.008 g/mol for H. The molar mass of LiOH is 23.95 + 15.999 + 1.008 = 40.957 g/mol. Therefore, the percent composition of Li in LiOH is (23.95/40.957) x 100 = 58.44%.
The molar mass of aluminum nitrate (Al(NO3)3) is 213 g/mol. The molar mass of nitrogen in the compound is 14 g/mol. Therefore, the percentage by mass of nitrogen in aluminum nitrate is (3 * 14) / 213 * 100 ≈ 19.7%.
15g Li * (1mol Li / 6.941g Li) = 2.16 mol Li
The element with the least mass in Period 2 is Lithium (Li). It has an atomic mass of approximately 6.94 g/mol
To determine the mass of lithium required to react with nitrogen gas, we first need to calculate the amount of nitrogen gas present using the ideal gas law. Then, we can use the stoichiometry of the balanced chemical equation between lithium and nitrogen to find the mass of lithium needed for complete reaction according to the molar ratio in the balanced equation.
7 X 102 g Al this is the awnser using significant figures
5.0 mol Li * 6.941 g/mol Li = 34.705 = 35 g Lithium