0.25moles
50 mg is .05 gm
2 NaOH + MgSO4-----Mg(OH)2+Na2SO4 1 MOLE OF MgSO4 produce one mole of Mg(OH)2 SO 1 MWT OF MgSO4 PRODUCE1MWT OF Mg(OH)2 HENCE 120 GM MgSO4 PRODUCE 58 gm OF Mg(OH)2 and so on if x gm of MgSO4 PRODUCE y gm OF Mg(OH)2 x=(120*y)/58 ---- ----
60 grams is 0.06kg
28.35 gm in 1 ounce. So 550 gm is 19.4 ounces
To prepare a 1% solution of sodium citrate, you would mix 1 gram of sodium citrate with 99 grams of water (for a total of 100 grams solution). Stir the mixture until the sodium citrate is fully dissolved in the water.
C2H4O2 + NaOH = H2O + C2H3O2Na Acetic acid (60 gm) + sodium hydroxide ( 40 gm) = 100 gm water (18 gm) + sodium acetate (82 gm) = 100 gm Ratio reactants to products = 1:1 Molarity = moles / L, 3M = 3 moles / 1 L Acetic acid = 60 gm / total reactant 100gm = 1.8 moles Multiply by 3 = 1.8 moles or 180 grams Sodium Hydroxide = 40 gm / total reactant 100 mg = 1.2 moles or 120 grams. 180 grams acetic acid + 120 grams sodium hydroxide = 300 grams. 300 grams divided by 1 liter = 3M So in order to make 3 M sodium acetate combine solution, add 180 grams acetic acid and 120 grams sodium hydroxide with 1 liter of water.
9.991 Moles (water) 8.982 Moles (heavy water)
The molar mass of barium (Ba) is approximately 137.3 g/mol. So, the mass of 3 moles of barium would be 3 moles x 137.3 g/mol = 411.9 grams.
The atomic weight of magnesium is 24.31; therefore, the number of moles in 100 gm is 100/24.31 = 4.11, to the justified number of significant digits.
According to wikipedia the density of sodium is 0.968 gm.cm-3. This is a mass of 0.160 gm. Then this much sodium would occupy 0.160 gm / ( 0.968 gm.cm-3 ) = 0.165 cm3.
Since sodium chloride has equal parts of sodium and chlorine by weight, you would need 29.3 grams of sodium to create 29.3 grams of sodium chloride.
50 mg is .05 gm
density of sodium citrate is 1.7 gm/ml for pure solid, whereas varies according to the concentration
To calculate the number of moles in 76 grams of NaCl, you need to divide the given mass by the molar mass of NaCl. The molar mass of NaCl is approximately 58.5 g/mol. So, 76 grams of NaCl is equal to 76 g / 58.5 g/mol ≈ 1.30 moles.
To find the number of moles in 16 g of oxygen gas, you need to divide the given mass by the molar mass of oxygen. The molar mass of oxygen (O2) is 32 g/mol. Therefore, 16 g / 32 g/mol = 0.5 moles of oxygen gas.
AgNo3(aq) + NaCl(aq) -------------->AgCl (ppt) + No3(-) + Na(+) no silver nitrates are produced it is all consumed. only silver chloride is produced and precipitate . free nitrate and free sodium ions are produced but do not react with each other 1 Mole AgNo3 ------->169.9 gm 1 Mole Nacl ------->58.4 gm 1 Mole AgCl ------->143.3 gm 4.02 gm AgNO3 = (4.02 / 169.9) = 0.02366 M AgCl produced = 0.02366 M = 3.39 gm
1000grams in a kilogram.