The equation for the reaction is 4 Na + O2 -> 2 Na2O. This shows that, for complete reaction, one mole of oxygen is required for each four gram atomic masses of sodium. The gram atomic mass of sodium is 22.9898; therefore, 46 grams of sodium constitutes 2.00 moles of sodium, to more than the justified number of significant digits. The gram molecular mass of diatomic oxygen is 31.9988; therefore 160 grams of oxygen constitutes 5.000 moles of diatomic oxygen, to more than the justified number of significant digits. This is well over the minimum amount of oxygen required for complete reaction of all the sodium present. Each two gram atomic masses of sodium produces one gram formula mass of sodium oxide; therefore, the number of gram formula masses of sodium oxide produced is 1.00, to at least the justified number of significant digits.
This amount may be different because rust is not a clearly definite compound.
160mm is 6.29921 inches. Direct Conversion Formula 160 mm* 1 in 25.4 mm = 6.299212598 in
160 kg = 352 lbs The formula to convert kg to lbs 160 kg*2.2046 lbs 1 kg=352.7396195 lbs
To find the current, use the formula: Power (W) = Voltage (V) x Current (A). Rearrange the formula to solve for current: Current (A) = Power (W) / Voltage (V). Therefore, 160 watts divided by 120 volts equals 1.33 Amps.
That is the chemical formula. The name would be iron (III) oxide, or ferric oxide in the old system.
To find the molecular formula, you need the empirical formula and molar mass. If the molar mass is 160 plus 5 grams per mole, the molecular formula cannot be determined without additional information about the empirical formula's molar mass relationship.
6g hydrogen would be required for 160g ferric oxide in this reaction. The relative atomic weights of the elements are: Hydrogen - 1 Oxygen - 16 Iron - 56 giving the relative atomic weights of the compounds (on the left of the equation): Fe2O3 = 56×2 + 16×3 = 160 3H2 = 3×(1×2) = 6 So for every 160 units of mass of iron III oxide there will be 6 units of mass of hydrogen required. → for 160g of iron III oxide ÷ 160 × 6 = 6 g of hydrogen.
This question seems to be about the reactant side of a chemical equation. To calculate the mass of oxygen produced, you need to know the stoichiometry of the reaction. Without that information, it is not possible to determine why 160 grams of oxygen would be produced.
The balanced reaction equation is 2Fe2O3 + 3C = 4Fe + 3CO2 Note the molar ratios are 2:3::4:2 Next calculate the acutal moles of Fe2O3 First using the Periodic Table Calculate the Mr ( Relative molecular mass) of Fe2O3 Fe x 2 = 56 x 2 = 112 O x 3 = 16 x 3 = 48 112 + 48 = 160 mol(Fe2O3) = 16.5 / 160 = 0.103125 Equivlanet to ;2; moles. mol(C) = 0.103125 x 3/2 = 0.1546875 equivalent to '3 moles. Mass(C) = 0.154687 x 12 = 1.85625 g (The Answer!!!!)
160...cant quite grasp HOW though
Good one. Fe2O3 is 160g/mol, so %comp of Fe is 112/160=0.7 or 70%. 1x106g Fe/70=x/30, so x=428,571g O2, so mass of Fe2O3 is 1,428,571g. 1,428,571/78=x/22, so x=402,930g of other rock, so mass of ore deposit = 1,428,571+402,930=1,831,501g=1,831.5kg.
The equation for the reaction is 4 Na + O2 -> 2 Na2O. This shows that, for complete reaction, one mole of oxygen is required for each four gram atomic masses of sodium. The gram atomic mass of sodium is 22.9898; therefore, 46 grams of sodium constitutes 2.00 moles of sodium, to more than the justified number of significant digits. The gram molecular mass of diatomic oxygen is 31.9988; therefore 160 grams of oxygen constitutes 5.000 moles of diatomic oxygen, to more than the justified number of significant digits. This is well over the minimum amount of oxygen required for complete reaction of all the sodium present. Each two gram atomic masses of sodium produces one gram formula mass of sodium oxide; therefore, the number of gram formula masses of sodium oxide produced is 1.00, to at least the justified number of significant digits.
To find the remaining mass of a radioactive isotope after a certain time, you can use the radioactive decay formula: [M_{\text{final}} = M_{\text{initial}} \times \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}] Given that the half-life of (^{222}\text{Rn}) is 3.8 days, and the initial mass is 160 milligrams, you can substitute these values into the formula to find the final mass.
V (VOLUME) = M (MASS) / D (DENSITY) All you have to do is to find the density of methylated spirit and use the formula
This amount may be different because rust is not a clearly definite compound.
The percentage can be calculated using the formula (152/160) * 100 = 95%. Therefore, 152 is 95% of 160.