i need help i have a homework assignment to finish and this is the last question help please and thank u

PLEASE HELP ME SOLVE THIS.Thank you so much!

Tju [1.3M]

Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

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

Now, the two half-reactions can be written as follows.

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

This will be balanced as follows.

$Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$ ... (1)

Oxidation half-reaction: $Cl^{-} \rightarrow Cl_{2} + 1e^{-}$

This will be balanced as follows.

$6Cl^{-} \rightarrow 3Cl_{2} + 6e^{-}$ ... (2)

Adding both equation (1) and (2) we will get the resulting equation as follows.

$Cr_{2}O^{2-}_{7} + 14H^{+} + 6Cl^{-} \rightarrow 2Cr^{3+} + 3Cl_{2} + 7H_{2}O$

Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.

6 0

3 years ago

Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

Katen [24]

Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0

3 years ago

Balance the following equation in acidic conditions Phases are optional. S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

jonny [76]

To balance the the chemical reaction, the number of moles per element is balance is both side of the reaction and also the charge in both sides of the reation. to balnce the reaction:

S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+

2S2O3 2- + Cu 2+ ---> S4O6 2- + Cu+ + e

3 0

3 years ago

CH3-CH2-CH=CH-CH=CH2

o-na [289]

Answer:

Hydration (of an alkene)

Mechanism : Electrophilic addition.

8 0

3 years ago

I need help!! i need to make it like 3:2

anyanavicka [17]

Answer:

use visual studio code and put in this print{3:2}-1

Explanation:

4 0

3 years ago