Mole percent of O2 = 10% = 0.1 moles
Mole percent of N2 = 10% = 0.1 moles
Mole percent of He = 80% = 0.8 moles
Molar Mass of O2 = (2 x 16) x 0.1 = 3.2
Molar Mass of N2 = (2 x 14) x 0.1 = 2.8
Molar Mass of He = 4 x 0.8 = 3.2
1. Molar Mass of the mixture = 3.2 + 2.8 + 3.2 = 9.2 grams
2. Since at constant volume density is proportional to mass, so the ratio of
mass will be the ratio of density.
Ratio = Molar Mass of the mixture / Molar Mass of O2 = 9.2 / 32 = 0.2875
The answer is A. Nitrogen Oxide
<u>Answer:</u> The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane
<u>Explanation:</u>
The IUPAC nomenclature of alkanes are given as follows:
- Select the longest possible carbon chain.
- For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
- A suffix '-ane' is added at the end of the name.
- If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.
We are given:
An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms
The image of the given alkane is shown in the image below.
Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane
Answer:
It donates a hydrogen ion
Explanation:
Under the Bronsted-Lowry definition of an acid, acids are protons donors which donate the H+ ion, or the hydrogen ion.
Answer: First, here is the balanced reaction: 2C4H10 + 13O2 ===> 8CO2 + 10H2O.
This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.
Next, let's determine how many moles of butane are burned. This is obtained by
5.50 g / 58.1 g/mole = 0.0947 moles butane. As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947 = 0.1894 moles CO2.
Now we need to figure out the volume. This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions: 273 K and 1 atmosphere.
We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).
V = 0.1894 x 0.082 x 273.0 / 1 = 4.24 Liters.
Explanation:
