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irakobra [83]
2 years ago
11

Study the following unbalanced half-reaction. Which equation represents the balanced half-reaction? H2O2 ---> H2O

Chemistry
2 answers:
MA_775_DIABLO [31]2 years ago
7 0

Answer : The correct option is, 2e^-+2H^++H_2O_2\rightarrow 2H_2O

Explanation :

The given unbalanced reaction is,

H_2O_2\rightarrow H_2O

(1) The right side is lacking oxygen, so we add a water molecule to it :

H_2O_2\rightarrow H_2O+H_2O

or we can write as,

H_2O_2\rightarrow 2H_2O

(2) Now we have introduced more hydrogen, so we need to balance hydrogen on the left side in acidic solution, we get

2H^++H_2O_2\rightarrow 2H_2O

(3) Now all the atoms are balanced but charge is not balanced. We see that the left side there are two positive charge. So, we balance the charge by adding two negative charge on the left side in the reaction, we get

2e^-+2H^++H_2O_2\rightarrow 2H_2O

Therefore, the correct option is, 2e^-+2H^++H_2O_2\rightarrow 2H_2O

Aleonysh [2.5K]2 years ago
5 0
O: 1*2 = 2*1 
<span>H 2 + 2 = 2*2 </span>

<span>answer C hope you get it right</span>
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One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori
Igoryamba

Answer:

P_2=0.398bar=39800Pa

T_2=118.7K\\

Q=-3729.9J

W=-61753.24J

ΔU_T=0J

ΔH_T=0J

Explanation:

Hello,

At the first state, the molar volume is:

v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3

The volume in both the second and third state:

v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3

Now, as it is about an adiabatic process, one remembers the following relationships:

PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa

- And the temperature:

T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J

Then the total heat:

Q=Q_1+Q_2=0-3729.9J=-3729.9J

- The work for the first process is:

W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

W=W_1+W_2=-61753.24J+0J=-61753.24J

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\

Best regards.

8 0
2 years ago
Select the word or phrase from the drop-down menu to describe ionic compounds. A formula unit represents the simplest ratio of e
Gelneren [198K]

Answer:

CRYSTAL

MANY ATOMS THAT ARE ARRANGE IN A REGULAR PATTERN

1:1

Explanation:

8 0
2 years ago
Read 2 more answers
A container of hydrogen at 172 kPa was decreased to 85.0 kPa producing a new volume of 3L. What was the original volume?
Aneli [31]

Answer:

<h2>The answer is 1.48 L</h2>

Explanation:

In order to find the original volume we use the same for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the original volume

V_1 =  \frac{P_2V_2}{P_1}  \\

From the question

P1 = 172 kPa = 172000 Pa

P2 = 85 kPa = 85000 Pa

V2 = 3 L

We have

V_1 =  \frac{85000 \times 3}{172000}  =  \frac{255000}{172000}  =  \frac{255}{172}  \\  = 1.482558139...

We have the final answer as

<h3>1.48 L</h3>

Hope this helps you

4 0
2 years ago
The temperature of 100. grams of water changes from 16.0ºC to 20.0ºC. What is the total number of Joules of heat energy absorbed
mafiozo [28]

<span>1.    </span>Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
Heat = 100(4.184)(20 - 16)
<span>Heat = 1673.6 J</span></span>

5 0
2 years ago
Read 2 more answers
Identify the gas that has a root mean square velocity of 412 m/s at 191 K (potassium)
AVprozaik [17]
Gas pump is the main reason 177m 192 k
5 0
2 years ago
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