Its described as a Straight Branch, hope this helps :)
1) Acidic solution. An acidic solution has a high concentration of H+ ions.
2) Basic solution. A basic solution has a high concentration of OH- ions.
3) HCl chemical reaction.
You can also write it as follows
4) pH scale.
When the concentration of H+ is high, the solution is acidic. H+ lowers the pH of the solution.
Option D.
We can determine the empirical formula by first converting each of the grams to moles. remember to do this, first, we need the molar mass of the molecules which can be calculated by adding the mass of the atoms from the periodic table.
molar mass of CO2= 44.0 g/mol
molar mass of H2O= 18.02 g/mol
now, lets determine the grams of each atom
Carbon: 23.98 g x (12.011 g / 44.01 g) = 6.54 g C
Hydrogen: 4.91 g x (2.0158 g / 18.02 g) = 0.55 g H
Oxygen: 10.0 - (6.54 + 0.55) = 2.91 g O
Now let's convert each mass to moles.
C: 6.54 g / 12.01 g / mol = 0.54 mol
H: 0.55 g / 1.01 g/mol = 0.54 mol
O: 2.91 g / 16.00 g/mol = 0.18 mol
now that we have the moles of each atom, we need to divide them by the smallest value to find the ration. If you do not get the whole number, you need to multiply until to get a whole number.
C: 0.54 mol / 0.18 mol = 3
H: 0.54 mol / 0.18 mol = 3
O: 0.18 mol / 0.18 mol = 1
empirical formula--> C₃H₃O
Answer:
Percent yield of CO₂ is 6.25 %.
Explanation:
Given data:
Percent yield of CO₂ = ?
Actual yield of CO₂ = 28.16 g
Number of moles of C₈H₁₈ = 8.000 mol
Number of moles of O₂ = 16.00 mol
Solution:
Chemical equation:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
Now we will compare the moles of C₈H₁₈ and O₂ with CO₂.
C₈H₁₈ : CO₂
2 : 16
8.000 : 16/2×8.000 = 64 mol
O₂ : CO₂
25 : 16
16 : 16/25×16= 10.24 mol
Less number of moles of CO₂ are produced from 16 moles of O₂. it will limit the yield of CO₂.
Grams of CO₂ produced:
Mass = number of moles × molar mass
Mass = 10.24 mol × 44 g/mol
Mass = 450.56 g
Percentage yield of CO₂:
Percentage yield = actual yield / theoretical yield × 100
Percentage yield = 28.16 g/ 450.56 g× 100
Percentage yield = 6.25 %
