Answer:
ΔHrxn = 239 kj/mol
Explanation:
Given data:
Mass of sodium = 0.230 g
Heat produced = 2390 J
Solution:
Chemical equation:
2Na + 2HCl → 2NaCl + H₂
Number of moles of sodium:
Number of moles of sodium = mass / molar mass
Number of moles of sodium = 0.230 g / 23 g/mol
Number of moles of sodium = 0.01 mol
Enthalpy of reaction:
ΔHrxn = 2390 J / 0.01 mol
ΔHrxn = 239000 j/mol
ΔHrxn = 239 kj/mol
Answer:
[H3O+] = 1.4*10^-5 M
pH = 4.85
[OH-] = 7.08*10^-10
Explanation:
As pH is a measure of hydronium H3O concentration, simply substitute [H3O+] into the following equation:
pH = -log[H+]
pH = -log(1.4*10^-5)
pH = 4.853871...
Round to 2-3 sig figs due to only being given data with 2 significant figures
pH = 4.85
One method to get to [OH-] is to turn pH into pOH and then use inverse functions to get [OH-]
pH + pOH = 14
4.85 + pOH = 14
pOH = 9.15
Then to get [OH-] from pOH:
pOH = -log[OH-]
9.15 = -log[OH-]
-9.15 = log[OH-]
10^(-9.15) = [OH-]
7.07945784 * 10^-10 = [OH]
Round based on given significant figures again:
7.08*10^-10 = [OH-]
(Feel free to add any questions & I'll be sure to reply if clarification is needed)
Answer:
.14L or 140mL
Explanation:
This is a classic plug-n-chug problem. Your textbook probably goes over this formula as . M stands for molarity of the given substance, and V stands for the volume that the substance occupies.
Simply plug in the values that you're given, like so:
After completing the algebra portion and solving for the unknown, you will be left with x = .14L, which is the volume required to neutralize 30mL of 7M NaOH.
The answer is D. Wind
Wind is considered physical weathering