To find the Percent Composition of an atom, you use this formula:
Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent.
So we get:
So you get about:
So, the percent composition of Hydrogen in NH4HCO3 is 6.3%
Hope this Helps! :D
Answer:
<h2>Percentage error is a measurement of the discrepancy between an observed and a true, or accepted value .</h2>
Explanation:
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Answer:
Según el científico inglés John Dalton, los átomos son esferas elásticas e indivisibles. Así, según él, el átomo es el bloque de construcción más pequeño de la materia. Es homogéneo e indivisible, y todos los átomos de un elemento químico dado son idénticos (es decir, tienen el mismo conjunto de propiedades).
Aunque se descubrió a finales del siglo XIX que los átomos están hechos de partículas aún más pequeñas y pueden sufrir transformaciones, y que los átomos de un elemento dado pueden diferir ligeramente entre sí (isótopos), la teoría de Dalton fue la base para el desarrollo de la tecnología química moderna.
<u>Answer:</u> The molality of 
solution is 0.782 m
<u>Explanation:</u>
Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molality:
.....(1)
Given values:
Moles of = 0.395 mol
Mass of solvent (water) = 0.505 kg
Putting values in equation 1, we get:
Hence, the molality of solution is 0.782 m
Answer:
A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:
C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol
The value given for ΔH∘rxnΔHrxn∘ means that:
a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.
b. the reaction is endothermic.
c. the enthalpy of formation of propane is 2202 kJ/mol.
d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.
e. None of these.
