The partial pressure of Hydrogen gas can directly be calculated
by simply taking the difference of the overall pressure and the vapour pressure
of water. That is:
P (H2 gas) = 759.2 torr – 23.8 torr
<span>P (H2 gas) = 735.4 torr</span>
Covalent Bond.
To be specific, it is polar covalent bond. :)
<u>Answer:</u> The solubility of oxygen at 682 torr is 
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:

Or,

where,
are the initial concentration and partial pressure of oxygen gas
are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr

Putting values in above equation, we get:

Hence, the solubility of oxygen gas at 628 torr is 
Answer:
0.48 V
Explanation:
Zn(s) ------------> Zn^2+(aq) + 2e. Oxidation half equation (-0.76V)
Co^2+(aq) + 2e-----------> Co(s). Reduction half equation (-0.28)
Zn(s) + Co^2+(aq) -------------> Zn^2+(aq) + Co(s) overall redox equation
Zinc is the anode while cobalt is the cathode.
E°cell= E°cathode - E°anode
E°cell= -0.28-(-0.76)= 0.48 V
Answer:
Actual yield = 86.5g
Explanation:
Percent yield = 82.38%
Theoretical yield = 105g
Actual yield = x
Equation of reaction,
CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
Percentage yield = (actual yield / theoretical yield) * 100
82.38% = actual yield / theoretical yield
82.38 / 100 = x / 105
Cross multiply and make x the subject of formula
X = (105 * 82.38) / 100
X = 86.499g
X = 86.5g
Actual yield of CaCl₂ is 86.5g