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Answer:
Explanation:
We want to convert from moles of water to grams of water.
First, find the molar mass of water (H₂O) Look on the Periodic Table for the masses of hydrogen and oxygen.
- Hydrogen (H): 1.008 g/mol
- Oxygen (O): 15.999 g/mol
Next, add up the number of each element in water. The subscript of 2 comes after the H, so there are 2 moles of hydrogen.
- 2 Hydrogen: (1.008 g/mol*2) = 2.016 g/mol
Finally, add the molar mass of 2 hydrogen and 1 oxygen.
- 2.016 g/mol (2 Hydrogen) + 15.999 g/mol (1 oxygen)= 18.015 g/mol
Next, find the grams in 6.5 moles.
Use the molar mass we just found as a ratio.
We want to find the grams in 6.5 moles. We can multiply the ratio above by 6.5
Multiply. Note that the moles of H₂O will cancel each other out.
If we want to round to the technically correct significant figures, it would be 2 sig figs. The original measurement, 6.5, has 2 (6 and 5).
<h3>0.020 × 1000 × 100</h3>
<h2>= 2000 mg of Sn</h2>
hope that helps !
Temperature is a measure of the average kinetic energy of the particles in the sample. This is the statement that defines the temperature of a sample of matter.
The temperature of a system is defined simply as the average energy of microscopic motions of a single particle in the system per degree of freedom.
The microscopic motions in a solid matter is the principal vibrations of the constituent atoms about their sites. In an ideal monoatomic gas, the microscopic motions are the translational motions of the constituent gas particles. In multiatomic gases, aside from translational motions, vibrational and rotational motions are included in the microscopic motions.
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
