The 3rd option is the answer
Depends on the height of the mountain. At 2500km (7500ft) it might boil-albeit reluctantly-at 100 degrees Celsius, but higher up not so much.
<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:
Density is a physical quantity, defined as the ratio of body mass to the volume occupied by this body. The average body density is the ratio of body weight to its volume.
Since the mass in a body can be distributed unevenly, a more adequate model defines the density at each point of the body as a derivative of mass over volume.
Thus, to obtain the density of a sample, its mass must be divided by its volume. Thus, the density of the sample is 1.2 / 1.1, that is, 1.09 g/cm3.
Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.
Explanation:
- d = m/V, where d is the density, m is the mass and V is the volume.
- We have the mass m = 0.50 g, so we must get the volume V.
- To get the volume of a gas, we apply the general gas law PV = nRT
P is the pressure in atm (P = 1.5 atm)
V is the volume in L (V = ??? L)
n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).
- Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
- Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.
