Answer:
Empirical formula: BH3
Molecular Formula: B2H6
Explanation:
To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:
100% _____ 27 g
78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron
100% ______27 g
21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen
100% _____ 28 g
78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron
100% _____ 28g
21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen
So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.
The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.
The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.
Answer: Predominant tropical easterly winds sweep across the central and southern portions of the state, keeping the temperatures mild
Explanation:
bc
Given:
0.103 grams of Zinc (solid)
50 ml of HCl
Density of HCl = 1.0 g/mL
Initial Temperature (Ti) = 22.5 C
Final Temperature (Tf) = 23.7 C
Solve for the mass of HCl:
50 mL * 1.0 g/mL = 50 g
Assume that the Cp for HCl is similar to the Cp of water:
q = mCpdT
= mCp (Tf - Ti)
= 50 g * 4.18 J/gC * (23.7 - 22.5)
q = 250.8 J = H
The Answer is A CH2O
The molecular formula for glucose is C6H12O6. The subscripts represent a multiple of an empirical formula. To determine the empirical formula, divide the subscripts by the GCF of 6 which gives CH2O.
