<u>Answer:</u> The osmotic pressure is 54307.94 Torr.
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
where,
= osmotic pressure of the solution = ?
i = Van't hoff factor = 3
C = concentration of solute = 0.958 M
R = Gas constant =
T = temperature of the solution =
Putting values in above equation, we get:
Hence, the osmotic pressure is 54307.94 Torr.
Answer:
B. 4.52 X10-9 M
Explanation:
Our goal for this question is to <u>calculate the concentration of hydronium ions</u> produced by water in a vessel with a concentration of hydroxide ions of . So, our first approach can be the <u>ionization reaction of water</u>:
If we write the <u>Keq expression</u> for this reaction we will have:
Now, water is the universal solvent, so, Keq has a special name. In the equilibrium problems for water we have to use <u>"Kw" instead of "Keq"</u>:
From this equation, we know the Kw value () and the concentration of the hydroxide ions ([2.21X10^-^6~M]). If we replace these values into the equation we can solve for :
I hope it helps!
Answer:
3.52 grams is the theoretical yield of carbon dioxide formed from the reaction of of methane and of oxygen gas.
Explanation:
Moles of methane gas =
Moles of oxygen gas =
According to reaction 1 moles of methane gas reacts with 2 mole of oxygen gas.
Then 0.08 moles of methane will recast with:
oxygen gas
As we can see that moles of methane are in limiting amount , so the amount moles of carbon dioxide will depend upon on moles of methane gas.
Limiting reagent = methane
According to reaction , 1 mole of methane gas gives 1 mol of carbon dioxide gas.
Then 0.08 moles of metahjen gas wuill give :
of carbon dioxide gas.
Mass of 0.08 mople sof carbon dioxde gas :
3.52 grams is the theoretical yield of carbon dioxide formed from the reaction of of methane and of oxygen gas.
Answer:
37.1 g of Arsenic
Explanation:
2.98 x 10^21 atoms of Arsenic x 74.92g/6.02 x 10^23
= 223.26 x 10^21/6.02 x 10^23
= 2.2326 x 10^23/6.02 x 10^23
= 37.1 g of Arsenic
Answer:
a) the minimun of acetic anhydride required for the reaction is 2.175 g (CH3CO)2O
b) V acetic anhydride = 2.010 mL
Explanation:
C6H4OHCOOH + (CH3CO)2O ↔ C9H8O4 + C2H4O2
⇒ mol salicylic acid = 2.94 g C6H4OHCOOH * ( mol C6H4OHCOOH / 138.121 g ) = 0.0213 mol C6H4OHCOOH
⇒ mol acetic anhydride = 0.0213 mol C6H4OHCOOH * ( mol (CH3CO)2O / mol C6H4OHCOOH ) = 0.0213 mol (CHECO)2O
⇒ g acetic anhydride = 0.0213 mol * ( 102.1 g/mol ) = 2.175 g CH3CO)2O
b) V = 2.175 g (CH3CO)2 * ( mL / 1.082 g ) = 2.010 mL (CH3CO)2O