Answer:
It should be silicon... I even googled it to see if I was correct and that’s what it says so can someone figure out why silicon is not an answer
Explanation:
Answer: The statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
Explanation:
A strong acid upon dissociation gives a weak conjugate base. This can also be said as stronger is the acid, weaker will be its conjugate base or vice-versa.
Hydrofluoric acid is a strong base as it dissociates completely when dissolved in water.
For example, ![HF \rightleftharpoons H^{+} + F^{-}](https://tex.z-dn.net/?f=HF%20%5Crightleftharpoons%20H%5E%7B%2B%7D%20%2B%20F%5E%7B-%7D)
The conjugate base is
which is a weak base.
Acetic acid is a weak acid as it dissociates partially when dissolved in water. So, the conjugate base of acetic acid is a strong base.
![CH_{3}COOH \rightarrow CH_{3}COO^{-} + H^{+}](https://tex.z-dn.net/?f=CH_%7B3%7DCOOH%20%5Crightarrow%20CH_%7B3%7DCOO%5E%7B-%7D%20%2B%20H%5E%7B%2B%7D)
Thus, we can conclude that the statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
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:
equal to
Explanation:
This question is depicting the LAW OF CONSERVATION OF MASS/MATTER, which states that matter can neither be created nor destroyed. Hence, in an isolated system, the original mass of the reactants must be EQUAL TO the final mass of the products formed.
In other words, the law states that no matter the change in chemical composition of the reactants in a chemical process, the mass does not change or it remains constant.
<u>Answer:</u>
<em>The molarity of the
solution is
</em>
<em></em>
<u>Explanation:</u>
The Balanced chemical equation is
![1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)](https://tex.z-dn.net/?f=1AgNO_3%20%28aq%29%20%2B1KCl%20%28aq%29%20%3E%201%20AgCl%20%28s%29%2B1KNO_3%20%28aq%29)
Mole ratio of
: KCl is 1 : 1
So moles
= moles KCl
![Moles KCl = \frac {mass}{molarmass}](https://tex.z-dn.net/?f=Moles%20KCl%20%3D%20%5Cfrac%20%7Bmass%7D%7Bmolarmass%7D)
![= \frac {0.785 mg}{(39.1+35.5 g per mol)}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7B0.785%20mg%7D%7B%2839.1%2B35.5%20g%20per%20mol%29%7D)
![= \frac {0.000785 g}{74.6 g per mol}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7B0.000785%20g%7D%7B74.6%20g%20%20per%20mol%7D)
![= 0. 0000105 mol KCl](https://tex.z-dn.net/?f=%3D%200.%200000105%20mol%20KCl)
![= 0.0000105 mol AgNO_3](https://tex.z-dn.net/?f=%3D%200.0000105%20mol%20AgNO_3)
So Molarity
![= \frac {moles of solute}{(volume of solution in L)}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7Bmoles%20of%20solute%7D%7B%28volume%20of%20solution%20in%20L%29%7D)
![= \frac {0.0000105 mol}{26.2 mL}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%20%7B0.0000105%20mol%7D%7B26.2%20mL%7D)
= 0.000402M or mol/L is the Answer
(Or)
is the Answer