2H = 2.016 grams
1S = 32.07 grams
4O = 64 grams
-------------------------------add
= 98.086 grams/mole
===============
Multiply the number of atoms with the molacular weights.
2*1 + 32 + 16*4 = 98 g/mol
or more accurately: 2*1.007947 + 32.065 + 4*15.99943 = 98.078614 g/mol
2 * [(2*1.008) + 32.06 + (4*16.00)] = 2 * 98.079 = 196.2 g
(1*2)+(32*1)+(16*4)
2+32+64
98
VOLUME OF water
(2 x 1) + 32 + (4 x 16) = 98
2
ITS .5
1 mole H2SO4 = 98.078g H2SO4
1 mole H2SO4 x 4 moles O/mole H2SO4 x 6.02x10^23 atoms of O/mole O = 2.4x10^34 oxygen atoms
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
1 gm mole of Sulphuric Acid (H2SO4) weights 98 gm.
Quite a few! 335 moles H2SO4 (4 moles O/1 mole H2SO4)(6.022 X 1023/1 mole O) = 8.07 X 1026 atoms of oxygen =======================
1 mole H2SO4 = 98.078g H2SO4
1 mole H2SO4 x 4 moles O/mole H2SO4 x 6.02x10^23 atoms of O/mole O = 2.4x10^34 oxygen atoms
Grams NaOH?? Balanced equation. 2NaOH + H2SO4 --> Na2SO4 + 2H2O 4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH) = 4.0 grams NaOH needed =================
1 gm mole of Sulphuric Acid (H2SO4) weights 98 gm.
Quite a few! 335 moles H2SO4 (4 moles O/1 mole H2SO4)(6.022 X 1023/1 mole O) = 8.07 X 1026 atoms of oxygen =======================
17.9325 grams H2SO4 (1 mole H2SO4/98.086 grams(4 moles O/1 mole H2SO4)(6.022 X 1023/1 mole O) = 4.40387 X 1023 atoms of oxygen ======================
1 mole Since H2SO4's molar mass is 98.1g, and a substance's molar mass contains exactly 1 mole, that is the answer
0.1 N means you have a hydronium ion concentration of 0.1 mole/L. Since sulfuric acid is a diprotic acid, you get 2 moles of ionizable protons per mole a sulfuric acid. Meaning, you only need a concentration of 0.05 M sulfuric acid to obtain 0.1 N sulfuric acid. [10 L] * [0.1 mole H+/L] * [1 mole H2SO4/2 mole H+] * [98.08 g/mole H2SO4] * [1 mL/1.84 g H2SO4] = 26.65 mL H2SO4 required to make 10 L 0.1 N H2SO4.
196 grams H2SO4 (1 mole H2SO4/98.096 grams)(6.022 X 10^23/1 mole H2SO4) = 1.20 X 10^24 molecules of sulfuric acid
2 moles H2SO4 (98.086 grams/1 mole H2SO4)= 196.172 grams of sulfuric acid====================
About 36 N, 36 mole H+/L. (It contains 18 mole/L H2SO4)
Go ahead and write the balanced equation: H2SO4 + 2NaOH --> Na2SO4 +2H2O If you have 1 mole of the reactant H2SO4, it will yield 2 moles of the product water. You can tell by the coefficients.