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irakobra [83]
3 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]3 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]3 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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3 years ago
A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and<br> 32 C.
lisabon 2012 [21]

Answer:

1255.4L

Explanation:

Given parameters:

P₁  = 928kpa

T₁  = 129°C

V₁  = 569L

P₂ = 319kpa

T₂  = 32°C

Unknown:

V₂  = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

           \frac{P_{1} V_{1} }{T_{1} }   = \frac{P_{2} V_{2} }{T_{2} }

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

 take the units to the appropriate ones;

             kpa to atm,  °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁  = 928kpa gives 9.16atm

T₂  = 32°C gives 273 + 32  = 305K

T₁  = 129°C gives 129 + 273  = 402K

Input the values in the equation and solve for V₂;

        \frac{9.16  x 569}{402}   = \frac{3.15 x V_{2} }{305}

       V₂   = 1255.4L

4 0
3 years ago
A student mixes a 10.0 ml sample of 1.0 m naoh(aq) with a 10.0 ml sample of 1.0 m hcl(aq) in a polystyrene container. the temper
AleksAgata [21]
Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
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             = (10 + 10) (4.184) (26-20) = 502.08 J
Q reaction = - Q water = -502.08 J
Hrxn = -502.08 / (0.01) = - 50208 J = - 50.21 kJ/mol
5 0
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