4.01 grams of Licl are needed to release this amount of heat
<h3>What does one gram weigh?</h3>
The gram is a unit of mass in the International System of Units (SI) that is equal to one thousandth of a kilogram. It was originally known as the gramme. Gram. This pen cap weighs around one gram. A weight scale like this one may provide a precise mass readout for many different things.
Mass of Licl required to release 5850J of heat to the surroundings.
Let x g Licl required.
Then x g ×∆H = 5850J
x g × 1.46 × 10³ = 5850
x = 5850/ 1.46 × 10³
x = 4.0068
so 4.01g Licl required.
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Answer:
None of these
Explanation:
For a reaction;
aA + bB ------>cC + dD
The equilibrium constant K is given as;
K = [C]^c [D]^d/[A]^a [B]^b
The equilibrium constant neither depends on the concentrations of the reactants nor on that of the products.
Let us recall that at equilibrium, the concentrations of reactants and products remain largely constant. This implies that, concentration of species do not appreciably change at equilibrium because the rates of forward and reverse reactions are equal.
Hence, the equilibrium constant neither depends on the initial/final concentrations of the reactants nor on the initial/final concentrations of the products.
Responda:
+ 0,9kJ / mol
Explicação:
Dados os calores de combustão do enxofre monoclínico e enxofre rômbico como - 297,2 kJ / mol e - 296,8 kJ / mol, respectivamente para a variação na transformação de 1 mol de enxofre rômbico em enxofre monoclínico conforme mostrado pela equação;
S (mon.) + O2 (g) -> SO2 (g)
Uma vez que são todos 1 mol cada, a mudança na entalpia será expressa como ∆H = ∆H2-∆H1
Dado ∆H2 = -296,8kJ / mol
∆H1 = -297,2kJ / mol
∆H = -296,8 - (- 297,2)
∆H = -296,8 + 297,2
∆H = 297,2-296,8
∆H = + 0,9kJ / mol
Portanto, a mudança na entalpia da equação é + 0,9kJ / mol
Hydrogen Carbonate, or HCO3.
