The correct answer is C. because it could lead to an increase in ocean levels
If heat is transferred from an object to the surroundings, then the object can cool down and the surroundings can warm up. When heat is transferred to an object by its surroundings, then the object can warm up and the surroundings can cool down.
A. 8,330 --> 8.33 × 10^3
B. 83.3 --> 8.33 × 10^1
C. 0.00833 --> 8.33 × 10^-3
D. 0.0833 --> 8.33 × 10^-2
E. 83,330 --> 8.33 × 10^4
Hope this helps!
Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
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<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
Answer:
The van't Hoff factor is 2.55
Explanation:
Step 1: Data given
Osmotic pressure of CaCl2 = 0.585 atm
Osmotic pressure of urea = 0.237 atm
Concentration = 9.69 * 10^-3 M
Temperature = 25.0 °C
Step 2:
Π = iMRT
⇒ with Π = the osmotic pressure of CaCl2 = 0.585 atm
⇒ with i is the van't Hoff factor
⇒ with M = the molar concentration = 9.69 * 10^-3 M
⇒ with R = the gas constant = 0.0821 L*atm/mol*K
⇒ with T = the temperature = 25.0 °C = 298 K
0.605 atm = i(9.69 * 10^-3 M)(0.0821 L*atm/mol*K)(298K)
i = 2.55
The van't Hoff factor is 2.55