To find the mass of 4.2 moles of FeO2, first calculate the molar mass of FeO2. Iron (Fe) has a molar mass of approximately 55.85 g/mol, and oxygen (O) has a molar mass of about 16.00 g/mol. Therefore, the molar mass of FeO2 is 55.85 g/mol + (2 × 16.00 g/mol) = 87.85 g/mol. Multiplying the molar mass by the number of moles gives: 4.2 moles × 87.85 g/mol = 369.07 grams.
As a rule of thumb, the atomic mass of an element equals the number of grams of that element equals a mole. Since the atomic mass of Nitrogen is 14, there are 14 grams in one mole of Nitrogen. Next, we just have to divide 42 by 14 and we get our answer: There are 2.9988 moles in 42 grams of Nitrogen.
1 kilogram = 1,000 grams 42 kilograms = 42,000 grams
Density = (mass) / (volume) = 2.33 gram/cc(rounded)Note: The 'ml' and the 'cc' are identical volumes.
One kilogram = 1000 grams Therefore 42g = 42 divided by 1000 = 0.042kg
To find the mass of phosphorus pentafluoride (PF₅) that has the same number of fluorine atoms as 25.0 g of oxygen difluoride (OF₂), we first determine the number of fluorine atoms in OF₂. Each molecule of OF₂ contains 2 fluorine atoms, and the molar mass of OF₂ is approximately 42 g/mol, meaning 25.0 g of OF₂ contains about 1.19 moles (25.0 g / 42 g/mol). Since there are 2 fluorine atoms per molecule of OF₂, this equates to approximately 2.38 moles of fluorine atoms. PF₅ contains 5 fluorine atoms per molecule, so to match 2.38 moles of fluorine atoms, we need 0.476 moles of PF₅ (2.38 moles F / 5). The molar mass of PF₅ is about 130 g/mol, so the mass of PF₅ needed is approximately 61.9 g (0.476 moles × 130 g/mol).
To calculate the number of moles in 42 grams of CO2, we first need to find the molar mass of CO2, which is approximately 44 grams per mole. Then, divide the given mass by the molar mass to find the number of moles. In this case, 42 grams of CO2 is equivalent to 42/44 = 0.955 moles.
To find the number of moles of Na in 42 grams, we can use the molar mass of Na, which is approximately 23 grams/mol. First, calculate the number of moles by dividing the given mass by the molar mass: 42 grams / 23 grams/mol = 1.83 moles of Na.
As a rule of thumb, the atomic mass of an element equals the number of grams of that element equals a mole. Since the atomic mass of Nitrogen is 14, there are 14 grams in one mole of Nitrogen. Next, we just have to divide 42 by 14 and we get our answer: There are 2.9988 moles in 42 grams of Nitrogen.
The molar mass of sodium hydrogen carbonate is 84 grams per mole, therefore 0.5 moles of it weighs 42 grams.
To calculate the number of moles in 25 grams of NaCl, divide the given mass by the molar mass of NaCl. The molar mass of NaCl is approximately 58.44 g/mol. Therefore, 25 grams of NaCl is equivalent to approximately 0.43 moles.
moles = mass/molar mass The molar mass of an oxygen atom = 16 g mol-1, as there are two oxygen atoms in diatomic oxygen this has to be doubled. 42g / 32g mol-1 = 1.3125 moles
To find the number of moles in 42 g of NaCl, divide the given mass by the molar mass of NaCl. The molar mass of NaCl is approximately 58.5 g/mol (sodium's atomic weight is 23 and chlorine's is 35.5). So, 42 g / 58.5 g/mol ≈ 0.72 moles of NaCl.
Density = (mass) / (volume) = 42/28 =1.5 gm/cm3
nominally 42 grams
That is aluminum hydroxide and as a polyatomic ion hydroxide needs to be in parenthesis. Thus; Al(OH)3 This shows aluminum's 3+ oxidation state and the three matching hydroxides 1- state. 42 grams Al(OH)3 ( 1 mole Al(OH)3/78.004 grams) = 0.538 moles Al(OH)3 As significant figures require; 0.54 moles Al(OH)3
This is not a valid conversion; milligrams (mg) and grams (g) are measures of weight or mass and mL (milliliters) is a measure of volume.
First you need to write a balanced equation. You are given that propane undergoes a combustion reaction that produces carbon dioxide and water.Unbalanced: C3H8 + O2 ---> CO2 + H2OBalanced: C3H8 + 5O2 ---> 3CO2 + 4H2OGivens:42.0 grams C3H8 (Molecular mass 44.0 g)115.0 grams O2 (Molecular mass 32.0 g)Molecular mass of CO2: 44.0 gMole ratio 1:5:3:4 (C3H8:O2:CO2:H2O)Then you need to find which of the reactants are the limiting reactant (lowest value) and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.42.0 g C3H8 / (44.0 g) = .955 moles C3H8 115.0 g