Answer:
The reason for the suspicion was because the manner in which iodine reacted chemically as well as its other chemical properties, indicated that it belonged in the same group as chlorine and bromine, while the much heavier tellurium should be placed in the previous group
The suspicion was proved to be correct when the atomic number of tellurium was found to be 52 and that of iodine was found to be 53 by later scientists
Explanation:
Answer:
0.0305 moles of MgCO₃
Explanation:
In order to solve this problem, we first need to calculate the molecular weight of MgCO₃:
- MgCO₃ MW = Atomic mass Mg + Atomic mass C + (Atomic mass O)*3
- MgCO₃ MW = 24.3 + 12 + 16*3 = 84.3 g/mol
Finally we <u>divide the mass by the molecular weight</u>, to calculate the <em>number of moles</em>:
- 2.57 g MgCO₃ ÷ 84.3 g/mol = 0.0305 moles.
The answer is D. Most common semiconducting materials are crystalline solids. A<span>morphous and liquid semiconductors are also known to be.</span>
Answer:
0.1 m C6H12O6 < 0.1 m NaCl < 0.1 m Cu(NO3)2 < 0.1 m Fe(NO3)3
Explanation:
Step 1: Data given
ΔTb = Kb*m*i
⇒with Kb = 0.512°C/m
⇒with m = molality
⇒with i = Van't hoff factor = the number of particles into which the solute dissociates
a. 0.1 m NaCl
⇒ molality = 0.1 molal
⇒ i(NaCl) = 2 ( dissociates in Na+ and Cl-)
0.1 * 2 = 0.2
b. 0.1 m Cu(NO3)2
⇒ molality = 0.1 molal
⇒ i(Cu(NO3)2) = 3 ( dissociates in Cu^2+ and 2NO3-)
0.1 * 3 = 0.3
c. 0.1 m C6H12O6
⇒ molality = 0.1 molal
⇒ i(C6H12O6) = 1 ( doesn't dissociate in water)
0.1 * 1 = 0.1
d. 0.1 m Fe(NO3)3
⇒ molality = 0.1 molal
⇒ i(Fe(NO3)3) = 4 ( dissociates in Fe^3+ and 3NO3-)
0.1 * 4 = 0.4
0.1 m C6H12O6 < 0.1 m NaCl < 0.1 m Cu(NO3)2 < 0.1 m Fe(NO3)3
The proton has two adjacent protons, so it splits into three peaks, a triplet. The proton has three adjacent protons, so it splits into four peaks, a quartet.
