Here is the solution,
for 1mole NH3 or ammonia,
N: 14.0067 g/mol x 1
H: 1.00794 g/mol x 3
So approximately it is--> 17.03052 g/mol)
Now multiply this into 4.5 for 4.5 moles of ammonia, --> 4.5x17.03052 =76.64g
45/94.2 is 0.4777 moles
To determine how many grams of zinc chloride (ZnCl₂) are needed to react with 45 grams of aluminum (Al), we first need to look at the balanced chemical equation for the reaction, which is: [ 2Al + 3ZnCl₂ \rightarrow 2AlCl₃ + 3Zn ] From the equation, 2 moles of aluminum react with 3 moles of zinc chloride. The molar mass of aluminum is approximately 27 g/mol, so 45 grams of aluminum corresponds to about 1.67 moles of Al. Thus, we need 2.5 moles of ZnCl₂ (since 1.67 moles Al will require 2.5 moles ZnCl₂). The molar mass of ZnCl₂ is about 136.3 g/mol, so 2.5 moles of ZnCl₂ would weigh approximately 340.75 grams.
To find the number of molecules in 45 grams of aluminum trifluoride (AlF₃), we first need to calculate the molar mass of AlF₃, which is approximately 26.98 g/mol (Al) + 3 × 19.00 g/mol (F) = 83.98 g/mol. Next, we can determine the number of moles in 45 grams by dividing the mass by the molar mass: 45 g / 83.98 g/mol ≈ 0.536 moles. Finally, using Avogadro's number (approximately (6.022 \times 10^{23}) molecules/mol), we find the number of molecules: 0.536 moles × (6.022 \times 10^{23}) molecules/mol ≈ (3.22 \times 10^{23}) molecules of AlF₃.
That is 45 teaspoons.
1.5873283 ouncesHere is the formula in case you are interested -45 g*1 oz28.34952313 g=1.587328288 oz
45/94.2 is 0.4777 moles
To determine how many grams of zinc chloride (ZnCl₂) are needed to react with 45 grams of aluminum (Al), we first need to look at the balanced chemical equation for the reaction, which is: [ 2Al + 3ZnCl₂ \rightarrow 2AlCl₃ + 3Zn ] From the equation, 2 moles of aluminum react with 3 moles of zinc chloride. The molar mass of aluminum is approximately 27 g/mol, so 45 grams of aluminum corresponds to about 1.67 moles of Al. Thus, we need 2.5 moles of ZnCl₂ (since 1.67 moles Al will require 2.5 moles ZnCl₂). The molar mass of ZnCl₂ is about 136.3 g/mol, so 2.5 moles of ZnCl₂ would weigh approximately 340.75 grams.
To calculate the number of grams in 2.50 moles of H2O, you first need to know the molar mass of water (H2O), which is approximately 18.015 g/mol. Then, you can use the formula: grams = moles x molar mass. So, 2.50 moles of H2O is equivalent to 45.04 grams (2.50 moles x 18.015 g/mol).
The chemical equation for the combustion of glucose is C6H12O6 + 6O2 -> 6CO2 + 6H2O. From this, we can see that 1 mole of glucose produces 6 moles of CO2. Therefore, 45 grams of glucose (which is approximately 0.25 moles) would produce 6 times that amount of CO2, which is about 90 grams.
C6H12O6 + 6O2 --> 6CO2 + 6H2O 45 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles CO2/1 mole C6H12O6)(44.01 grams/1 mole CO2) = 66 grams carbon dioxide produced ==========================
There's a thousand grams in a kilogram, 45 grams is 0.045 kilograms.
0.045 grams is 45 milligrams.
450 grams
First of all we convert the mass of leadto moles. Using the equation moles = mass(g) / Ar (Realtive atomic Mass)_. Mass( g) = 45 g Ar =~ 207 (Periodic Table). Hence moles(Pb) = 45 x 207 Moles (Pb) = 9315 moles. Next using the Avogadro number. 1 moles(of A SUBSTANCE) contains 6.022 x 10^(23) atoms. Hence number of atoms in 9315 moles = 9315 x 6.022 x 10^(23) = 5.61 x 10^(27) atoms. As a 'silly' number. 5,610,000,000,000,000,000,000,000,000 atoms.
1kg = 1000g. Therefore, 45 x 1000 = 45000 grams.
To find the number of molecules in 45 grams of aluminum trifluoride (AlF₃), we first need to calculate the molar mass of AlF₃, which is approximately 26.98 g/mol (Al) + 3 × 19.00 g/mol (F) = 83.98 g/mol. Next, we can determine the number of moles in 45 grams by dividing the mass by the molar mass: 45 g / 83.98 g/mol ≈ 0.536 moles. Finally, using Avogadro's number (approximately (6.022 \times 10^{23}) molecules/mol), we find the number of molecules: 0.536 moles × (6.022 \times 10^{23}) molecules/mol ≈ (3.22 \times 10^{23}) molecules of AlF₃.
6.25 kg/45 litres = 6250 grams/45 litres = 6250/45 grams per litre =138.88... gpl