Answer:
HF + H₂O ⇔ H₃O⁺ + F⁻
Explanation:
Bronsted- Lowery acid:
Bronsted- lowery acid is the specie which have capability to donate the proton.
Bronsted- Lowery conjugate base:
The Bronsted- Lowery conjugate base of Bronsted- Lowery acid is the specie which is formed after when an acid donate the proton.
Bronsted- Lowery base:
Bronsted- lowery base is the specie which have capability to accept the proton.
Bronsted- Lowery conjugate acid:
The Bronsted- Lowery conjugate acid of Bronsted- Lowery base is the specie which is formed after when an base accept the proton.
Chemical equation:
HF + H₂O ⇔ H₃O⁺ + F⁻
In this equation HF is acted as Bronsted- Lowery acid because it donate the proton to water. water is act as Bronsted- Lowery base. While F⁻ is the Bronsted- Lowery base and H₃O⁺ Bronsted- Lowery acid.
In a series circuit, the current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component.
Answer:
13.5 * 10^-2 g
Explanation:
What we know:
Balanced Equation: 3Ba+Al2(SO4)3 -->2Al+3BaSO4,
Grams of Ba: 1
Grams of Al2(SO4)3: 1.8g
Calculate the # of moles of Ba and Al2(SO4)3:
1g Ba/137.3 = 7.3 *10^-3 mol Ba
1.8g Al2(SO4)3/ 342 = 5.3 *10^-3 mol Al2(SO4)3
Find the limiting reactant:
Ba has a coefficient of 3 in the balanced equation, so we divide the # of moles of Ba by 3 to get... 7.3 *10^-3 mol Ba/3 = 2.43 *10^-3
Al2(SO4)3 has a coefficient of 1, so if we divide by 1, we get the same number of 5.3 *10^-3
2.43 *10^-3 is smaller than 5.3 *10^-3, therefore Ba is the limiting reactant.
finally, we just find the number of moles of Al
The ratio of Al to Ba is 2:3 so...
7.3 * 10^-3 * (2/3) = 5 *10^-3 mol Al
CONVERT TO GRAMS
5 *10^-3 mol Al * 27 = 13.5 * 10^-2 g
<u>Hope that was helpful! </u>
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
<span>A theory which repeatedly fails to confirm the expected predictions probably should be discarded. If the given theory is failing once or twice, then the facts and figures can be researched much more to understand the real cause of the error caused. But in case, the theory is again and again fails to confirms the expected predictions, then there is a gap in the facts and figures and the theory being studied and hence it should be discarded.</span>