<span>6.03 moles.
1 molecule of butane contains 4 carbon atoms and ten hydrogen atoms.
The molar mass is 4 times the atomic mass of carbon, 12 g/mol, plus 10 times the atomic weight of hydrogen, 1 g/mol.
Molar mass = 4 * 12 g/mol + 10 * 1 g/mol = 58 g/mol.
This means that 1 mole of butane has a mass of 58 g.
To figure out how many moles are in a sample of butane, divide the mass of sample in grams by 58 grams
Number of moles in sample = 350 g / 58 g/mol = 6.03 moles.</span>
<span>Answer:
A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?
--------------------------------------...
Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different.
25.305% C/12 = 2.108
74.695% Cl/35.5 = 2.104
So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.)
0.044 grams/10 ml = x/22.4 liters
0.044g/0.010 liters = x/22.4 liters
22.4 liters/0.010 liters = 2240 (ratio)
2240 x .044 = 98.56 (actual atomic weight)
CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole.
This is sufficiient to distinguish C2CL2, (dichloroacetylene)
from C6CL6 (hexachlorobenzene) which would
mass 3 times as much.</span>
Answer:I think it’s material chemistry
Explanation:
<span>It is known
that acids compounds contains hydrogen and produces hydrogen ion in water. A binary
acid however is an acid that have two elements, one of the element has a
hydrogen attached to it. Examples of binary acids are hydrogen fluoride (HF),
hydrogen bromide (HBr) and hydrogen sulfide (H2S). In naming a binary acid, it
has two rules; one, as pure compounds and two, as acid solutions. For pure
compounds, start with the name ‘hydrogen’ and end the anion name with ‘-ide’. For
acidic compounds, start with ‘hydro-‘, end the anion with ‘-ic’ and add ‘acid’.</span>