Answer:
– 1
Explanation:
From the question given above, we obtained the following:
Electron = 8
Net charge of nitrogen =.?
Nitrogen has atomic number of 7. This also means that nitrogen has 7 proton because atomic number of an element is the equal to number of protons in the atom of the element.
Thus, we can obtain the net charge of nitrogen with 8 electrons by calculating the difference between the protons and electrons of the nitrogen atom. This can be obtained as follow:
Proton = 7
Electron = 8
Net charge = Proton – Electron
Net charge = 7 – 8
Net charge = – 1
Therefore, the net charge of the nitrogen atom with 8 electrons is – 1
Answer: 1.027 x 10^6 g= 1027kg
In this question, you are given the volume of the blimp (2.027×10^5 ft^3) and the density of the gas(0.179g/L). To answer this question, you need to convert the volume unit into liter. The calculation would be: 2.027×10^5 ft^3 x 28.3168L/ft3= 57.398 x 10^5L= 5.74x10^6L
Then to find the mass, multiply the volume with the density. The calculation would be: 5.74x10^6L x 0.179g/L= 1.027 x 10^6 g= 1027kg
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
Well this is a bio question but it goes the tunica interna the tunica media then the tunica externa.
