Answer:
Explanation:
Remark
This is one of those questions that you need the choices. You can't tell what you should enter. For example, in scientific notation, you would get 1.5*10^7.
Or you could keep it as an integer and round to two places 150,000,000,
I would pick scientific notation if you know how to use it. Otherwise use the interger format.
Your answer is B.
Because it says that that carbon burns in presence of oxygen (C+O) which is equal ( => ) to Carbon Dioxide (
)
CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3
Explanation:
Boiling point trend of Butane, Propan-1-ol and Propanal.
Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.
In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.
Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.
In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol
Answer: 0.4M
Explanation:
Given that,
Amount of moles of NaOH (n) = ?
Mass of NaOH in grams = 40.0g
For molar mass of NaOH, use the atomic masses: Na = 23g; O = 16g; H = 1g
NaOH = (23g + 16g + 1g)
= 40g/mol
Since, n = mass in grams / molar mass
n = 40.0g / 40.0g/mol
n = 1 mole
Volume of NaOH solution (v) = 2.5 L
Concentration of NaOH solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 1 mole / 2.5 L
c = 0.4 mol/L (Concentration in mol/L is the same as Molarity, M)
Thus, the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution is 0.4 mol/L or 0.4M
Answer:
the candle is still solid..................in this case, yes!
Explanation:
it is still solid because the molecules are packed together tighter than the molecules in a liquid or gas.
