Answer:
Limiting reactant: O2
grams NO2 produced = 230.276 g NO2
grams of NO unused = 26.67 gNO
Explanation:
2NO + O2 --> 2NO2
Step 1: Determine the molar ratio NO:O2
molar ratio NO:O2 = 5.895: 2.503 = 2.35
stoichiometric molar ratio NO:O2 = 2:1
So, O2 is the limiting reactant.
Step2: Determine the grams of NO2:
?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2
Step 3: Determine the amount of excess reagent unreacted
moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted
moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted
mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted
The concentration of diluted solution is 0.756M.
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 18.9 mL
Molarity of stock solution (M1) = 10 M
Volume of diluted solution (V2) = 250 mL
Molarity of diluted solution (M2) =?
We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:
M1V1 = M2V2
10 × 18.9 = M2 ×250
189 = M2 × 250
Divide both side by 100
M2 = 189 / 250
M2 = 0.756 M
Therefore, the molarity of the diluted solution is 0.756 M.
Thus the concluded that concentration of the dilute acid is 0.756 M.
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The potassium will donate one of its valence electrons
Answer:p-hydroxybenzaldehyde is stronger acid to phenol
para-cyanophenol is stronger acid to meta-cyanophenol
o-fluorophenol is stronger acid to p-fluorophenol.
Explanation:
The PKa tool relative to Ph are used to contrast the pairs.
The pKa of phenol is 10. The pKa of p-hydroxybenzaldehyde is 9.24
The pKa for meta-cyanophenol is 8.61 and the pKa for para-cyanophenol is 7.95.
The pKa value of o-fluorophenol is 8.7, while that of the p-fluorophenol is 9.9. It's obvious that the inductive effect is more dominant at ortho-position, which results in a more acidic nature
The pKa is the pH value at which a chemical species will accept or donate a proton. The lower the pKa, the stronger the acid and the greater the ability to donate a proton in aqueous solution.
374u
187u
C₁₄H₂₂N₄O₈
Explanation:
To find the molecular weight of the compound C₁₄H₂₂N₄O₈ we simply sum that atomic masses of the given elements in the compound.
The empirical weight is determined by using the simplest ratio of the elements involved in the compound;
Molecular weight of C₁₄H₂₂N₄O₈;
atomic mass of C = 12g/mol
H = 1g/mol
N = 14g/mol
O = 16g/mol
Molecular weight = 14(12) + 22(1) + 4(14) + 8(16)
= 168 + 22 + 56 + 128
= 374u
Empirical weight:
Empirical formula:
C₁₄ H₂₂ N₄ O₈
14 : 22 : 4 : 8
divide by 2:
7 : 11 : 2 : 4
empirical formula C₇H₁₁N₂O₄
empirical weight = 
= 
= 187u
The molecular formula is the actual combination of atoms in a compound. so the molecular formula of the compound is C₁₄H₂₂N₄O₈
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Molecular mass brainly.com/question/5546238
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