The EPA sets national air-quality standards for common air pollutants. The

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

$MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}$

Thus, we can conclude that half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

3 0

3 years ago

A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio

Pachacha [2.7K]

Answer:

$pK_{a}$ of HA is 6.80

Explanation:

$pK_{a}=-logK_{a}$

Acid dissociation constant ( $K_{a}$ ) of HA is represented as-

$K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

$K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}$

So, $K_{a}=1.6\times 10^{-7}$

Hence $pK_{a}=-log(1.6\times 10^{-7})=6.80$

6 0

3 years ago

Two scientists were competing to earn some grant money to fund their research, because

Citrus2011 [14]

Societal and economical

3 0

3 years ago

Read 2 more answers

"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the

soldier1979 [14.2K]

Answer: The specific heat of calorimeter is 30.68 J/g°C

Explanation:

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 46.7 g

$m_2$ = mass of cold water = 45.33 g

$T_{final}$ = final temperature = 59.4°C

$T_1$ = initial temperature of hot water = 80.6°C

$T_2$ = initial temperature of cold water = 40.6°C

$c_1$ = specific heat of hot water = 4.184 J/g°C

$c_2$ = specific heat of cold water = 4.184 J/g°C

$c_3$ = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

$46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)$

$c_3=30.68J/g^oC$

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0

3 years ago

What type of compound does the formula, CuCl2, represent? a. ionic salt b. covalent molecule

Rainbow [258]

The answer is: [B]: "ionic salt" .
___________________________________________________
Note: There is no "sharing of electrons" among the elements in this compound; so this compound in NOT a "covalent molecule".

However, there is ionic bonding: Cu²⁺ and Cl⁻ ; to form: "CuCl₂" .
____________________________________________________

7 0

3 years ago