Answer:
Explanation:
Well the gas is the fuel for the flame of course. The collision theory comes into play when the gas turns on, chemicals collide with one another. Then reactions occur causing the flame. Then when you take away the fuel, the flame stops because there is no atoms or molecules to come together/collide with one another.
Sorry if its wrong or doesn't make sense... Wish you the best of luck on whatever your doing!
5.7% KCl is 94.3 % water.
Therefore, for 1000 g of water the mass of KCl will be (1000× 5.7)/94.3 = 60.445 grams.
1 mole of KCl is equal to 74.55 g,
therefore, 60.445 g will be 60.445/74.55 = 0.8108 mole of KCl
Hence, 0.8108 moles of KCl should release twice that number of moles 1.6216 moles ions.
Having 1.6216 moles of KCl ions dissolved in 1000g of water, gives us 1.6216 molar if solution.
Using the freezing point depression constant of water.
dT = Kf (molarity)
dT = (1.86 C/ molar) (1.6216 m)
dT = 3.016 C drop in freezing point
Therefore, it should freeze at - 3.016 Celsius
Answer:
787.5 grams of glucose (C₆H₁₂O₆) can prepare 1750 mL of a 2.50 M solution
Explanation:
Molarity (M) is a concentration measure that indicates the number of moles of solute that are dissolved in a given volume.
Molarity is expressed in the following way:
Molarity is expressed in units .
2.50 M means that in 1 L of solution there are 2.5 moles of glucose. So, you apply a rule of three as follows: if in 1 L there are 2.5 moles of glucose, in 1.75 L (1750 mL, being 1000 mL = 1 L) how many moles of the compound are there?
moles=4.375
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of glucose is:
C₆H₁₂O₆= 6* 12 g/mole+ 12* 1 g/mole + 6* 16 g/mole
C₆H₁₂O₆= 180 g/mole
Then you can apply a rule of three as follows, knowing the moles in 1750 mL and the molar mass: if there are 180 g of glucose in 1 mole, how much mass is there in 4.375 moles?
mass= 787.5 g
<u><em>787.5 grams of glucose (C₆H₁₂O₆) can prepare 1750 mL of a 2.50 M solution</em></u>