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:
ionic or covalent
Explanation:
The outermost electrons -- the valence electrons -- are able to interact with other atoms, and, depending on how those electrons interact with other the atoms, either an ionic or covalent bond is formed, and the atoms fuse together to form a molecule.
Answer:
I got the answers but it won't let me post it correctly on here....
Explanation:
9.) 10-2.76 =0.0174 [H30+]= 1.74*10-3 M
10.)10-3.65=0.00224 [H3O+] =2.24*10-2 M
11.)10-3.65=0.00224 [OH-]= 2.224*10-4M
12.)10-6.87=0.00000135 [OH-]= 1.35*10-7M
Answer:
The molarity of the solution: 0,27M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl
A solution molar--> moles of solute in 1 L of solution:
4 L-----1,09 mol NaCl
1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M
