Answer:
x = 2+
Explanation:
1) FADH2 + Q => FAD + QH2
Since H is added to Q
=> Reactant reduced is Q
(2) Balancing charges on both sides of the equation gives:
QH2 + 2 cyt c(Fe3+) => Q + 2 cyt c(Fe2+) + 2 H+
Thus x = 2+
My main reason would be momentum – it depends on the mass as
well as the speed of the colliding objects. For example if two sedans
travelling in a low speed bump each other, then probably the damage would be
minimal scratches. However, if a high speed car crashes unto a heavy truck also
travelling fast, then the result would be catastrophic.
(57.0 g B2O3 / (69.6202 g B2O3/mol) x (4mol BCI3 / 2 mol B2O3) = 1.64 mol BC13
(44.7 g C12) / (70.9064 g C12/mol) x (4mol BCI3 / 6mol C12) = 0.42027 mol BC13
(68.8 g C) / (12.01078 G C/mol) x (4mol BCI3 / 3 mol C) = 7.63 mol BCI3
C12 is the limiting reactant.
(0.42027 mol BCI3) X (117 . 170 g BCI3/mol) = 49.2 g BCI3 in theory.
Answer:
We have to weigh 52.8 g of BaCl₂·2H₂O, add it to a 2.00 L flask and add water until reaching the final volume.
Explanation:
<em>Describe the preparation of 2.00 L of 0.108 M BaCl₂ from BaCl₂·2H₂O. (244.3 g/mol).</em>
Step 1: Calculate the moles of BaCl₂
We need to prepare 2.00 L of a solution that contains 0.108 moles of BaCl₂ per liter of solution.
2.00 L × 0.108 mol/L = 0.216 mol
Step 2: Calculate the moles of BaCl₂·2H₂O that contain 0.216 moles of BaCl₂
The molar ratio of BaCl₂·2H₂O to BaCl₂ is 1:1. The moles of BaCl₂·2H₂O required are 1/1 × 0.216 mol = 0.216 mol.
Step 3: Calculate the mass corresponding to 0.216 mol of BaCl₂·2H₂O
The molar mass of BaCl₂·2H₂O is 244.3 g/mol.
0.216 mol × 244.3 g/mol = 52.8 g
We have to weigh 52.8 g of BaCl₂·2H₂O, add it to a 2.00 L flask and add water until reaching the final volume.
