One lone pair, as there is a single covalent bond between the hydrogen and the carbon, and a triple covalent bond between the nitrogen and the carbon, leaving two free electrons/one electron pair on the nitrogen..... I think ((:
Initial observation is a nasty slime and a horrible odor.
(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: 0.0748 grams of CO will dissolve in 1.00 L of water if the partial pressure of CO is 2.75 atm
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant = 
= partial pressure of CO = 2.75 atm
Putting values in above equation, we get:
Hence, the solubility of carbon monoxide gas is 0.0748 g/L
Answer:
13
Explanation:
Let's consider the unbalanced equation for the combustion of propane.
C₃H₈ + O₂ ⇒ CO₂ + H₂O
We will begin balancing H atoms by multiplying H₂O by 4.
C₃H₈ + O₂ ⇒ CO₂ + 4 H₂O
Now, we balance C atoms by multiplying CO₂ by 3.
C₃H₈ + O₂ ⇒ 3 CO₂ + 4 H₂O
Finally, we get the balanced equation multiplying O₂ by 5.
C₃H₈ + 5 O₂ ⇒ 3 CO₂ + 4 H₂O
The sum of the coefficients is 1 + 5 + 3 + 4 = 13.
