Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
Answer:
a. Polar
b. Polar
c. Non-polar
d. Non-polar
Explanation:
a. 
, hydronium cation contains a positive charge. Just as any other ion, it is polar, as it has a net charge.
b. 
has the same shape as water. There are two lone pairs on sulfur atom which produce an overall dipole moment in this molecule, the bent structure is polar.
c. 
is non-polar, as the central atom, phosphorus, doesn't contain any lone pairs, all the dipole moments cancel out: two dipole moments in the vertical plane, P-Cl, and three P-Cl dipoles in the horizontal plane within a trigonal bipyramidal shape.
d. 
is non-polar, since it's a tetrahedral molecule with no lone pairs on carbon atom, all four C-F dipole moments cancel out to yield a net 0 dipole moment.
Answer:
B- The polarity of the molecules and hydrogen bonding between molecules.
Explanation:
Hope this helps:)
The pH of salt depends on the component acid and base that comprise them. For example, if the salt is made up of strong acid and weak base then, the salt is acidic. If the salt is formed from strong base and weak acid then, the salt is basic. For this item, NH4Cl is acidic and also Ca(NO3)2 is acidic.
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
