Answer:
Percentage Yield is calculated by doing -> experimental/theoretical x 100% = percentage yield.
I THINK that if your questions are connected it should be 14/16.89 x 100% = 82.889%
Explanation:
I believe it would be C. While D is also a good answer, the question asks for the statement that is most likely true, and there is no absolute guarantee that in 20 years our resource use will be higher than it is now.
Answer:
0.26×10²³ molecules
Explanation:
Given data:
Volume of gas = 1.264 L
Temperature = 168°C
Pressure = 946.6 torr
Number of molecules of gas = ?
Solution:
Temperature = 168°C (168+273= 441 K)
Pressure = 946.6 torr (946.6/760 = 1.25 atm)
Now we will determine the number of moles.
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1.25 atm ×1.264 L / 0.0821 atm.L/ mol.K ×441 K
n = 1.58 /36.21 /mol
n = 0.044 mol
Now we will calculate the number of molecules by using Avogadro number.
1 mol = 6.022×10²³ molecules
0.044 mol × 6.022×10²³ molecules/ 1mol
0.26×10²³ molecules
Answer:
6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺
Explanation:
Step 1: Write the unbalanced reaction
BrO₃⁻ + Sb³⁺ ⟶ Br⁻ + Sb⁵⁺
Step 2: Identify both half-reactions
Reduction: BrO₃⁻ ⟶ Br⁻
Oxidation: Sb³⁺ ⟶ Sb⁵⁺
Step 3: Perform the mass balance, adding H⁺ and H₂O where appropriate
6 H⁺ + BrO₃⁻ ⟶ Br⁻ + 3 H₂O
Sb³⁺ ⟶ Sb⁵⁺
Step 4: Perform the charge balance, adding electrons where appropriate
6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O
Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻
Step 5: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost is the same
1 × (6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O)
3 × (Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻)
Step 6: Add both half-reactions and cancel what is repeated in both sides
6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺