The answer is 0.25 g/cm3
The equation for finding density is mass divided by volume.
Therefore,
12 grams/ 49 cm3= 0.24489795918367 gram/cubic centimeter
you would then round this to the nearest significant figure which is the number of significant numbers you have and the highest amount of sig figs present in this equation is two sig figs. That means that you would round your answer, becasue it has a zero before the decimal which doesn't count as a significant figure, to the nearest hundredth which would be 0.25 and then you add your value after it which is g/cm3
Answer:
110.984 ?
i apologize if i'm wrong, you can report it if im wrong
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Explanation:
A. Chloroplasts
B. The cell wall and the vacuole
C. Vacuoles
D. The mitochondrion
You should do which candle burn the fastest colored or white
Given data:
Sublimation of K
K(s) ↔ K(g) ΔH(sub) = 89.0 kj/mol
Ionization energy for K
K(s) → K⁺ + e⁻ IE(K) = 419 Kj/mol
Electron affinity for Cl
Cl(g) + e⁻ → Cl⁻ EA(Cl) = -349 kj/mol
Bond energy for Cl₂
1/2Cl₂ (g) → Cl Bond energy = 243/2 = 121.5 kj/mol
Formation of KCl
K(s) + 1/2Cl₂(g) → KCl(s) ΔHf = -436.5 kJ/mol
<u>To determine:</u>
Lattice energy of KCl
K⁺(g) + Cl⁻(g) → KCl (s) U(KCl) = ?
<u>Explanation:</u>
The enthalpy of formation of KCl can be expressed in terms of the sum of all the above processes, i.e.
ΔHf(KCl) = U(KCl) + ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)
therefore:
U(KCl) = ΔHf(KCl) - [ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)]
= -436.5 - [89 + 419 + 243/2 -349] = -717 kJ/mol
Ans: the lattice energy of KCl = -717 kj/mol
