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Answer:
0.800 mol
Explanation:
We have the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with moles of the compounds involved.
Step 1. <em>Gather all the information</em> in one place.
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 4.00 4.00
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Step 2. Identify the <em>limiting reactant
</em>
Calculate the <em>moles of CO₂</em> we can obtain from each reactant.
<em>From C₃H₈:</em>
The molar ratio of CO₂: C₃H₈ is 3:1
Moles of CO₂ = 4.00 × 3/1
Moles of CO₂ = 12.0 mol CO₂
<em>From O₂</em>:
The molar ratio of CO₂: O₂ is 3:5.
Moles of CO₂ = 4.00 × ⅗
Moles of CO₂ = 2.40 mol CO₂
O₂ is the limiting reactant because it gives the smaller amount of CO₂.
==============
Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.
The molar ratio of C₃H₈:O₂ is 1:5.
Moles of C₃H₈ = 4.00 × ⅕
Moles of C₃H₈ = 0.800 mol C₃H₈
Answer:
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Explanation:
Using the formula
Ca Va = Cb Vb
Cb = 0.32 M
Vb = 50 mL = 50/1000 = 0.050L
Ca = 0.5 M
Va =?
Substituting for Va in the equation, we obtain:
Va = Cb Vb / Ca
Va = 0.32 * 0.05 / 0.5
Va = 0.016 / 0.5
Va = 0.032 L
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Delta E = Ef - Ei
E = energy , h = plank constant , v = frequency
h= 6.626 * 10 ^-34 j*s , T = 10 ^ 12 , v = 74 * 10 ^12 Hz , Hz = s^-1
E = ( 6.626 * 10^ -34 j*s) ( 74 * 10 ^ 12 s^ -1 ) = 4.90 * 10 ^ -20 J
Delta E = Ef - Ei
-4.90 * 10 ^ -20 J = -2.18 * 10 ^ -18J ( 1/4 ^2 - 1/x ^2)
0.0225 = 0.0625 - ( 1/x ^ 2)
0.225 - 0.0625 = - 1/ x ^ 2
- 0.0400 = - 1/x ^2 = -1 / - 0.0400 = x^2
25 = x^2
x = 5
Answer:
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