Answer:
6.1 cm³
Explanation:
To solve this problem we first need to keep in mind <em>Archimedes' principle</em>:
- The volume of water (or any fluid) displaced by a submerged object is equal to the object's volume.
With that in mind we <u>calculate the volume of the granite piece in mililiters</u>:
- Volume displaced = 47.6 mL - 41.5 mL = 6.1 mL
- Volume of the granite piece = 6.1 mL
Given that one cubic centimeter is equal to one mililiter, the volume of the granite piece in cm³ is 6.1 cm³.
Answer:
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Explanation:
When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:
P = iMRT,
for strong electrolytes, i = number of ions.
for nonelectrolytes, i = 1
1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa
The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa
<em>So, that means that NaCl is more hypertonic than the sucrose solution.</em>
2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
<em>In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.</em>
Answer:
-238.54 kJ/mol.
Explanation:
- We need to calculate the standard enthalpy of formation of liquid methanol, CH₃OH(l) that has the equation:
C(graphite) + 2H₂(g) + ½ O₂(g) → CH3OH(l) ΔHf° = ?
?? kJ/mol.
- using the information of the three equations:
(1) C(graphite) + O₂(g) → CO₂(g), ΔHf₁° = -393.5 kJ/mol
.
(2) H2(g) + ½ O₂(g) → H₂O(l), ΔHf₂° = -285.8 kJ/mol
.
(3) CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l), ΔH₃° = -726.56 kJ/mol
.
- We rearranging and add equations (1, 2, and 3) in such a way as to end up with the needed equation:
- equation (1) be as it is:
(1) C(graphite) + O₂(g) → CO₂(g), ΔHf₁° = -393.5 kJ/mol
.
- equation (2) should be multiplied by (2) and also the value of ΔHf₂°:
(2) 2H2(g) + O₂(g) → 2H₂O(l), ΔHf₂° = 2x(-285.8 kJ/mol
).
- equation (3) should be reversed and also the value of ΔH₃° should be multiplied by (-1):
(3) CO₂(g) + 2H₂O(l) → CH₃OH(l) + 3/2 O₂(g), ΔH₃° = 726.56 kJ/mol
.
- By summing the modified equations, we can get the needed equation and so:
The standard enthalpy of formation of liquid methanol, CH₃OH(l) = ΔHf₁° + 2(ΔHf₂°) - ΔH₃° = (-393.5 kJ/mol
) + 2(-285.8 kJ/mol
) - (- 726.56 kJ/mol) = -238.54 kJ/mol.
The relationship between volume and pressure is described in the Gay-Lussac's law. That is,
P₁V₁ = P₂V₂
Substituting the known values,
(75 mL)(145 atm) = (500 mL)(P₂)
The value of P₂ is equal to 21.75 atm.