Answer:
σ = 4.998 E-4 C/m²
Explanation:
- 1 Coulomb (C) ≡ 6.241509 E18 electrons (e)
∴ # elect = 6.24 E14 elect
charge (Q):
⇒ Q = (6.24 E14 elect)/( 1 C /6.241509 E18 elect) = 9.998 E-5 C
charge density (σ):
∴ surface area (S) = 0.2 m²
⇒ σ = ( 9.998 E-5 C ) / ( 0.2 m²)
⇒ σ = 4.998 E-4 C/m²
<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.
<u>Explanation:</u>
We are given:
39.0 % of sodium in sodium chloride solution
This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution
Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)
Applying unitary method:
If 39 grams of sodium metal is present in 100 grams of sodium chloride solution
So, if 0.1 grams of sodium metal will be present in = 
of sodium chloride solution.
Hence, the mass of sodium chloride solution present is 0.256 grams.
Answer:
42 19 K→42 20 Ca+e−
Explanation:
Naturally-occurring potassium atoms have a weighted average atomic mass of 39.10 (as seen on most modern versions of the periodic table.) Each potassium atom contains 19 protons p+ and thus an average potassium atom contains about 39.10−19≈20 neutrons n0.
This particular isotope of potassium-42 contains 42 nucleons (i.e., protons and neutrons, combined;) Like other isotopes of potassium 19 out of these nucleons are protons; the rest 42−19=23 are therefore neutrons.
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
