Answer:
16 mol NaCl.
Explanation:
Do the train track method to cancel out all the units except moles of NaCl on top. Remember one mole of any gas occupies 22.4 L at STP.
179.2 L CO2 x 1 mol CO2/22.4 L CO2 x 2 mol NaCl/1 mol CO2
= 16 mol NaCl
Answer:
The water would be neutral, (usually 7). The salt water would be the same (7) and the vinegar would be very acidic. (probably 2).
Explanation:
Answer:
2.445 g
Explanation:
Step 1: Given and required data
- Energy in the form of heat required to boil the water (Q): 5525 J
- Latent heat of vaporization of water (∆H°vap): 2260 J/g
Step 2: Calculate the mass of water
We will use the following expression.
Q = ∆H°vap × m
m = Q / ∆H°vap
m = 5525 J / (2260 J/g)
m = 2.445 g
Weight percentage of nitrogen can be calculated using the following rule:
weight percentage of nitrogen = (weight of nitrogen / weight of urea) x 100
From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of hydrogen = 1 grams
molecular mass of oxygen = 16 grams
therefore:
mass of nitrogen in urea = 2(14) = 28 grams
mass of urea = 12 + 2(14) + 4(1) + 16 = 60 grams
Substitute with the masses in the equation to get the percentage:
weight percentage of nitrogen = (28/60) x 100 = 46.667%
