I don’t see any equal signs to make it an equation. Am I missing something?
Answer:
The osmotic pressure is 8.85 atm
Explanation:
Step : Data given
Mass of glucose = 55.0 grams
Volume of water =800.0 mL
Temperature = 10.0 °C
Step 2: Calculate moles glucose
Moles glucose = mass glucose / molar mass glucose
Moles glucose = 55.0 grams / 180.156 g/mol
Moles glucose = 0.305 moles
Step 3: Calculate osmotic pressure
π = iMRT
⇒ with π = the osmotic pressure = TO BE DETERMINED
⇒ with i = the van't Hoff factor for glucose = 1
⇒ with M = the concentration = moles / volume = 0.305 moles / 0.800 L =0.381 M
⇒ with R = the gas constant = 0.08206 L*atm*mol¨K
⇒ with T = the temperature = 10.0 °C = 283 K
π = 1* 0.381 *0.08206 * 283
π = 8.85 atm
The osmotic pressure is 8.85 atm
It depends on the type of decay that's taking place, but if it's b+ it'd decay into Mn-52. If it's b- it'd decay into Co-59 (i'm sure??)
What is the question exactly?
63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
The balanced reaction equation is;
2 Na + 3 N2--> 2 NaN3
Number of moles of Na = 50.0 g/23 g/mol = 2.17 moles of Na
Number of moles of Nitrogen = 40.5 g/28 g/mol = 1.45 moles of N2
We have to obtain the limiting reactant, this is the reactant that yields the least number of moles of product.
For Na
2 moles of Na yields 2 moles of NaN3
2.17 moles of Na yields 2.17 moles of NaN3 (reaction is 1:1).
For N2
3 moles of N2 yields 2 moles of NaN3
1.45 moles of N2 yields 1.45 * 2/3 = 0.97 moles of NaN3
So, N2 is the limiting reactant. Mass of product formed depends on the limiting reactant.
Mass of NaN3 = 0.97 moles of NaN3 * 65 g/mol = 63.1 g of NaN3
Therefore, 63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
Learn more: brainly.com/question/9743981
