Explanation:
The given data is as follows.
= 286 kJ = 
= 286000 J
, 
Hence, formula to calculate entropy change of the reaction is as follows.
= ![[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20S_%7BO_%7B2%7D%7D%29%20-%20%281%20%5Ctimes%20S_%7BH_%7B2%7D%7D%29%5D%20-%20%5B1%20%5Ctimes%20S_%7BH_%7B2%7DO%7D%5D)
= ![[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20205%29%20%2B%20%281%20%5Ctimes%20131%29%5D%20-%20%5B%281%20%5Ctimes%2070%29%5D)
= 163.5 J/K
Therefore, formula to calculate electric work energy required is as follows.
= 
= 237.277 kJ
Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.
Answer:
a. 174 mL
Explanation:
Let's consider the following reaction.
2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)
We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:
0.1550 L × 0.112 mol/L = 0.0174 mol
The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:
2 × 0.0174 mol = 0.0348 mol
The volume of a 0.200 M KI solution that contains 0.0348 moles is:
0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL
Answer:
The empirical formula is the simplest form;
Given:
Oxygen O at 94.1% and
H at 5.9%
Assume 100grams.
94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O
5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H
There is one mole of O for each mole of H so the empirical formula is 
and written as OH.
1 mole of any substance contains 6.022 × 1023 particles.
⚛ 6.022 × 1023 is known as the Avogadro Number or Avogadro Constant and is given the symbol NA
N = n × NA
· N = number of particles in the substance
· n = amount of substance in moles (mol)
· NA = Avogardro Number = 6.022 × 10^23 particles mol-1
For H2O we have:
2 H at 1.0 each = 2.0 amu
1 O at 16.0 each = 16.0 amu
Total for H2O = 18.0 amu, or grams/mole
It takes 18 grams of H2O to obtain 1 mole, or 6.02 x 1023 molecules of water. Think about that before we answer the question. We have 25.0 grams of water, so we have more than one mole of water molecules. To find the exact number, divide the available mass (25.0g) by the molar mass (18.0g/mole). Watch how the units work out. The grams cancel and moles moves to the top, leaving moles of water. [g/(g/mole) = moles].
Here we have 25.0 g/(18.0g/mole) = 1.39 moles water (3 sig figs).
Multiply 1.39 moles times the definition of a mole to arrive at the actual number of water molecules:
1.39 (moles water) * 6.02 x 1023 molecules water/(mole water) = 8.36 x 1023 molecules water.
That's slightly above Avogadro's number, which is what we expected. Keeping the units in the calculations is annoying, I know, but it helps guide the operations and if you wind up with the unit desired, there is a good chance you've done the problem correctly.
N = n × (6.022 × 10^23)
1 grams H2O is equal to 0.055508435061792 mol.
Then 23 g of H2O is 1.2767 mol
To calculate the number of particles, N, in a substance:
N = n × NA
N = 1.2767 × (6.022 × 10^23)
N= 176.26
N=
