Answer:
When electrons are removed from an object, it becomes positively charged
Answer and Explanation:
The explanation given in the problem is correct but not totally encompassing.
Van der waals interactions are a type of hydrophobic interaction, in which they do not interact with the polar water molecule. Covalent bonds involve the sharing of electrons between atoms of relatively similar electronegativities, and are most often too strong to disrupt by polar molecules of water. Therefore, covalent bonds and van der waals forces have an Intrinsic bond strength value that is independent of the environment.
However, either the partial negative oxygen atom or the partial positive hydrogen atoms in water molecules disrupt hydrogen or ionic bonds. Water is known to form hydrogen bonds with other polar or charged molecules, thus reducing the strength of interaction these molecules would normally have in the absence of water. Basically, these compounds with Hydrogen or Ionic bonds ionize, whether partially or fully in water, thereby leading to a decrease in bond strength in water.
QED!
Answer:
C.
Explanation:
para sakin letter C ganonn
Answer : The final temperature would be, 791.1 K
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = 265 kJ/mol = 265000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B4%5Ctimes%20K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B265000J%2Fmol%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B733K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
Therefore, the final temperature would be, 791.1 K
Answer:
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l) + 6 e⁻
Explanation:
In order to balance a half-reaction we use the ion-electron method.
Step 1: Write the half-reaction
Cr³⁺(aq) = Cr₂O₇²⁻(aq)
Step 2: Perform the mass balance, adding H₂O(l) and OH⁻(aq) where appropriate
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l)
Step 3: Perform the electric balance, adding electrons where appropriate
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l) + 6 e⁻
