40 grams is 1.5 ounces 40 grams is 1.5 ounces
9 grams aluminum (1 mole Al/26.98 grams) = 0.3 moles aluminum ==============
3,56 mol Al is equal to 96,05 grams.
Quite a few! 698 grams Al(NO3)3 (1 mole Al(NO3)3/213.01 grams)(9 moles O/1 mole Al(NO3)3)(6.022 X 1023/1 mole O) = 1.78 X 1025 atoms of oxygen ======================
For this you need the atomic mass of Al. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel.2.91 moles Al × (27.0 grams) = 78.6 grams Al
2.38 mol Al
9 grams aluminum (1 mole Al/26.98 grams) = 0.3 moles aluminum ==============
بذر الكتان, حبْ الكتّان Habba Al-Kattan or Bazra Al-Kattan
3,56 mol Al is equal to 96,05 grams.
The molar weight of aluminum (Al) is 26.98 g/mol. grams of Al = 26.98 x 0.471 = 12.71 g The molar weight of aluminum (Al) is 26.98 g/mol. grams of Al = 26.98 x 0.471 = 12.71 g The molar weight of aluminum (Al) is 26.98 g/mol. grams of Al = 26.98 x 0.471 = 12.71 g
The mass of aluminium is 11,2 g.
10 grams aluminum (1 mole Al/26.98 grams) = 0.37 moles of aluminum ---------------------------------
To determine how many grams of aluminum hydroxide can be produced from 5.50 grams of aluminum sulfide (Al₂S₃), we first need to know the balanced chemical reaction involved. The reaction between aluminum sulfide and water produces aluminum hydroxide and hydrogen sulfide. The molar mass of Al₂S₃ is approximately 150.17 g/mol, and that of aluminum hydroxide (Al(OH)₃) is about 78.00 g/mol. Using stoichiometry, 1 mole of Al₂S₃ produces 2 moles of Al(OH)₃; thus, from 5.50 g of Al₂S₃, we can calculate that it can yield approximately 7.30 grams of aluminum hydroxide.
Quite a few! 698 grams Al(NO3)3 (1 mole Al(NO3)3/213.01 grams)(9 moles O/1 mole Al(NO3)3)(6.022 X 1023/1 mole O) = 1.78 X 1025 atoms of oxygen ======================
7 X 102 g Al this is the awnser using significant figures
Answer 14.2884 Grams 12 fluid oz. 4.2 % AL There are 28.35 grams in one ounce x 12 = 340.2 grams 340.2 x .042 = 14.2884 grams Roughly a 1/2 ounce
To find the mass of Cr formed, we first need to calculate the initial moles of Cr2O3 and Al. Next, we determine the limiting reactant based on stoichiometry. Knowing Al is the limiting reactant, we calculate the theoretical yield of Al2O3 formed, and then find the amount of Cr formed based on the stoichiometry of the reaction.
For this you need the atomic mass of Al. Take the number of grams and divide it by the atomic mass. Multiply it by one mole for units to cancel.94.5 grams Al / (27.0 grams) = 3.50 moles Al