Lar mass of Ca<span> = 40.08 </span>grams/mole 77.4 g Ca<span> * ( 1 </span>mole Ca<span>/ 40.08 ... n = m / M 1mol </span>Ca<span>weights 40 gmol-1 n = 77,4 / 40 = 1.93 </span>mol<span>.</span>
The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of </span>C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂
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Answer:
λ = 4.59 ×10⁻³ nm
Explanation:
Given data:
Energy of light = 4.33 ×10⁻¹⁴ J
Wavelength of light = ?
h = 6.626×10⁻³⁴ js
c = 3×10⁸ m/s
Solution:
Formula:
E = hc/λ
4.33 ×10⁻¹⁴ J = 6.626×10⁻³⁴ Js× 3×10⁸ m/s / λ
λ = 6.626×10⁻³⁴ Js× 3×10⁸ m/s / 4.33 ×10⁻¹⁴ J
λ = 19.878 ×10⁻²⁶ Jm / 4.33 ×10⁻¹⁴ J
λ = 4.59 ×10⁻¹² m
λ = 4.59 ×10⁻³ nm
