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.
The molar mass of sodium is 22.99 ㅤ ㅤ ㅤ 22.99 (28) = 643.72 mol now multiply by avogrados number to find the number of atoms. ㅤ ㅤ ㅤ 643.72 (6.022 x 10^23) = 3.88 x 10^26 number of atoms
Answer:
<h2>The answer is 35 mL</h2>
Explanation:
Density of a substance can be found by using the formula
From the question we are finding the amount of space the gas will occupy that's the volume of the gas
Making volume the subject we have
From the question
mass = 23.8 g/mL
Density = 0.68 g/mL
Substitute the values into the above formula and solve
That's
We have the final answer as
<h3>35 mL</h3>
Hope this helps you
Answer:
Bi (Bismuth)
Ag (Silver)
Li (Lithium)
Explanation:
Xe (Xenon) and I (Iodine) are non-metals. They cannot from a metallic bond because metallic bonds are bonds between metals only.
Convalent Bond energy is low
