The answer is 615.91 grams of <span>n2f4
Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams
pH of the solution after 24. 00 ml of the hcl has been added is 12.87
millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00
millimoles HCl = 24.00 mL x 0.10 M = 2.40
total volume = 48.00 mL
.................................NaOH + HCl ==>NaCl + H2O
initial.........................6.00.........0............0.........0
added.....................................2.40............................
change.................... -2.40......-2.40.........+2.40.... +2.40
equilibrium.................3.60.........0..............2.40.......2.40
The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = 0.075 M = (OH^-)
pOH = -log (OH^-). Then
pOH = -log (0.075)
pOH =1.1249
As we know,
pH + pOH = pKw = 14.00
pH=14-pOH
pH=14-1.1249
pH=12.87
<h3>
What is pH?</h3>
pH is a logarithmic measure of an aqueous solution's hydrogen ion concentration. pH = -log[H+], where log is the base 10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter.
The pH of an aqueous solution describes how acidic or basic it is, with a pH less than 7 being acidic and a pH greater than 7 being basic. A pH of 7 is regarded as neutral (e.g., pure water). pH values typically range from 0 to 14, though very strong acids may have a negative pH and very strong bases may have a pH greater than 14.
Learn more about pH:
brainly.com/question/491373
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I think it's the c (<span>.all the gases that surround the Earth)</span>
Answer: The energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.
Explanation:
Given: n = 2
The relation between energy and 
orbit of an atom is as follows.
Substitute the values into above formula as follows.
The negative sign indicates that energy is being released.
Thus, we can conclude that the energy of an electron in the n = 2 level of a hydrogen atom is 3.40 eV.
