False it was rutherford if i am not mistaken
Answer:
-3.82ºC is the freezing point of solution
Explanation:
We work with the Freezing point depression to solve the problem
ΔT = m . Kf . i
ΔT = Freezing point of pure solvent - freezing point of solution
Let's find out m, molality (moles of solute in 1kg of solvent)
15 g / 58.45 g/mol = 0.257 moles of NaCl
NaCl(s) → Na⁺ (aq) + Cl⁻(aq)
i = 2 (Van't Hoff factor, numbers of ions dissolved)
m = mol /kg → 0.257 mol / 0.250kg = 1.03 m
Kf = Cryoscopic constant → 1.86 ºC/m (pure, for water)
0ºC - Tºf = 1.03m . 1.86ºC/m . 2
Tºf = -3.82ºC
Answer:
c: Yes, if each sample contains 6.02 x 1023 atoms.
Explanation:
got it right on the test <3
<h2><u>Answer:</u></h2>
A nonpartisan iota of Nitrogen has a mass of 18. There are 7 protons in the core of this iota. What number of neutrons, complete electrons, and valence electrons are available
Nitrogen 15 has a nuclear mass of 15. The mass number is # protons in addition to # of neutrons, so for N-15 mass is 15 and the protons are dependably 7 so there must be 15-7=8 neutrons. N-15 has 7 electrons since it has 7 protons and p = e.
When 2.50 g is burned then in oxygen then 1.25kj of heat is produced.