<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 = ![30^oC=[30+273]K=303K](https://tex.z-dn.net/?f=30%5EoC%3D%5B30%2B273%5DK%3D303K)
Putting values in above equation, we get:

Hence, the osmotic pressure is 54307.94 Torr.
Answer:
Theoretical yield = 3.75g
Explanation:
Percent yield is defined as one hundred times the ratio between actual yield and theoretical yield. The expression is:
Percent yield = Actual Yield / Theoretical Yield * 100
In the problem, your actual yield was 3-00g.
Percent yield is 80.0%.
Solving for theoretical yield:
80% = 3.00g / Theoretical yield * 100
Theoretical yield = 3.00g / 80.0% * 100
<h3>Theoretical yield = 3.75g</h3>
I believe the correct response is A. Can be separated by physical means.
Answer:
2 NaBr(aq) + Cl₂(g) ⇒ 2 NaCl(aq) + Br₂(l)
Explanation:
Let's consider the unbalanced equation that occurs when aqueous sodium bromide reacts with chlorine gas to form aqueous sodium chloride and liquid bromine. This is a single displacement reaction.
NaBr(aq) + Cl₂(g) ⇒ NaCl(aq) + Br₂(l)
We will start balancing Cl atoms by multiplying NaCl by 2.
NaBr(aq) + Cl₂(g) ⇒ 2 NaCl(aq) + Br₂(l)
Then, we get the balanced equation by multiplying NaBr by 2.
2 NaBr(aq) + Cl₂(g) ⇒ 2 NaCl(aq) + Br₂(l)