There's 6.022×10^23 particles in 1 mole of anything
like there is 1000 grams in 1 kilogram of anything
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
The difference in an area with high concentration and an area with low concentration is called the concentration gradient.
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What is Concentration Gradient ?</h3>
A concentration gradient occurs when the concentration of particles is higher in one area than another.
In passive transport, particles will diffuse down a concentration gradient, from areas of higher concentration to areas of lower concentration, until they are evenly spaced.
This difference in an area with high concentration and an area with low concentration is called the concentration gradient.
Learn more about diffusion here ;
brainly.com/question/24746577
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Answer:
[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.
Explanation:
- For a weak acid like HF, the dissociation of HF will be:
<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>
[H₃O⁺] = [F⁻].
<em>∵ [H₃O⁺] = √Ka.C,</em>
Ka = 6.8 x 10⁻⁴, C = 0.710 M.
∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.
<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>
<em></em>
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>
Answer:
I'm thinking cooper but not sure
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