G mol to mol l
[DOC File]UNIT 3 – WORKSHEET 1: MOLE PROBLEMS
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Find the molecular formula of a compound if its empirical formula is CH2O and its molar mass is 90 g/mol. CH2O = 12 + (2 x 1) + 16 = 30 g/mol. 90/30 = 3, so empirical formula x 3 + molecular formula =m so MF = C3H6O3 _____ Title: UNIT 3 – WORKSHEET 1: MOLE PROBLEMS Author: Fairfax County Public Schools Last modified by ...
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(3) H2(g) + ½O2(g) ( H2O(l) (H = -285.8 kJ/mol In this lab you will experimentally determine the (H values for equations (1) and (2). It would be difficult for you to calculate the (H value for equation (3), so it is given (–285.8kJ/mol).
[DOC File]EQUILIBRIO QUIMICO
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26- Un recipiente cerrado contiene inicialmente 3,6 mol/l de N2O4(g) y 3,6 mol/l de H2O(g). La mezcla se calienta y los gases reaccionan hasta alcanzar el equilibrio: 2N2O4(g) + 6H2O(g) ( 4NH3(g) + 7O2(g) Entonces la concentración de vapor de agua es 0,6 mol/l. Calcula la concentración de O2 en el equilibrio. R: 3,5 mol/l
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mol C. 2 H 6 = 6 30 =0,2 mol; mol Cl 2 = 7,1 71 =0,1 mol dan mol C 2 H 5 Cl = 5,16 64,5 =0,08 mol C 2 H 6 + Cl 2 C 2 H 5 Cl mula-mula : 0,2 0,1
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70.90 g mol 22.4 mol L . 3.17 . g L (f) Ar39.95. g mol . 39.95 g mol 22.4 mol L . 1.78. g L . 5. What happens to the volume of a bicycle tire or a basketball when you use an air pump to add air? if n↑ then V↑ according to Avogadro's law. 6. Sometimes when you blow up a …
[DOCX File]Mayfield City Schools
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70.90 g mol 22.4 mol L . 3.17 . g L (f) Ar39.95. g mol . 39.95 g mol 22.4 mol L . 1.78. g L . 43. What happens to the volume of a bicycle tire or a basketball when you use an air pump to add air? if. n↑ then V↑ according to Avogadro's law. 44. Sometimes when you blow up a …
[DOC File]Molarity Notes – H
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Example 1: Molar mass of salt (NaCl) is 58.5 g/mol, so if you dissolve 58.5g of salt in enough water to make 1.00 liter of solution, you have one mole of solute dissolved per liter of solution. The concentration of this solution is one molar (1.00 mol/L or 1.00M)
[DOC File]BANCO DE QUESTÕES 2015
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Alternativa correta. A concentração 0,6 mg/L equivale a 1,75 x 10-6 mol/L. para efetuar esse cálculo, basta dividir a massa 0,6 x 10-3 g pela massa molar do sulfato de alumínio, que vale 342g/mol. 0,6 x 10-3g dividido por 342 g/mol é igual a 1,75 x 10-6 mol/L. Alternativa incorreta. Nesse caso o aluno errou ao trabalhar com potência de dez.
[DOC File]Chapter 13 worksheet #1
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Assume that you have 1 L of solution: Mass of water = mass of solution – mass of NaOH = 1000 g – (0.300 mol)(40 g/mol) = 1000 g – 12 g = 988 g. Molality = 0.300 mol/0.988 kg = 0.304 m. Notice that the molality is a little bigger than molarity. Why are these two numbers similar and why is molality always greater than molarity?
[DOCX File]AP Chemistry
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c.The molecular mass of the compound is approximately 300 g/mol. Determine the molecular formula. 24. 75.0 g of Fe reacts with 11.5 L of O 2 (d = 3.48 g/L) according to the reaction: 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) a.Calculate the initial moles of Fe(s) and O 2 (g).
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