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ra1l [238]
1 year ago
9

Sorry this is a bit long

Chemistry
1 answer:
Anestetic [448]1 year ago
8 0

The overall balanced reaction equation is;

4Zn(s) + 10H^+(aq) + NO3^-(aq) -----> 4Zn^2+(aq) + NH4^+(aq) + 3H2O(l)

<h3>What is the balanced reaction equation?</h3>

The redox reaction equation is said to be balanced when the number of electron gained is equal to the number of electrons lost.

Now;

1. Reduction and oxidation half-reactions

Zn(s) -----> Zn^2+(aq) + 2e

And

NO3^-(aq) ---->NH4^+(aq) + 3H2O(l)

2. Using the H2O and H+ to balance O and H;

4Zn(s) + 10H^+(aq) + NO3^-(aq) -----> 4Zn^2+(aq) + NH4^+(aq) + 3H2O(l)

3. Balancing the electrons lost and gained; 4Zn(s) + 10H^+(aq) + NO3^-(aq) + 8e -----> 4Zn^2+(aq) + NH4^+(aq) + 3H2O(l) + 8e

4. The overall balanced reaction equation is;

4Zn(s) + 10H^+(aq) + NO3^-(aq) -----> 4Zn^2+(aq) + NH4^+(aq) + 3H2O(l)

Learn more about redox reaction:brainly.com/question/13293425

#SPJ1

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3 years ago
Why steam distillation is usefull for thermally un stable compounds​
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Why does the temperature of the water stay the same when it melts and boils?
krok68 [10]

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5 0
3 years ago
What is the molality of a solution made by dissolving 15.20 g of i2 in 1.33 mol of diethyl ether, (ch3ch2)2o?
Paraphin [41]
The  molarity   of solution  made  by  dissolving  15.20g  of i2  in 1.33 mol  of diethyl ether (CH3CH2)2O  is    =0.6M

   calculation

molarity  =moles of solute/  Kg of the  solvent

mole  of the solute  (i2)  =  mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol

moles is therefore=  15.2 g/253.8 g/mol  =  0.06  moles


calculate the Kg of solvent  (CH3CH2)2O
mass =  moles  x  molar mass
molar mass  of  (CH3CH2)2O= 74 g/mol

mass  is therefore = 1.33 moles  x  74 g/mol =  98.42 grams
in Kg = 98.42 /1000 =0.09842  Kg

molarity  is therefore = 0.06/0.09842 = 0.6 M

3 0
3 years ago
You have 100 mL of a 12M solution of HCl, and you need to dilute it to 1.5M for an experiment. How many liters will your new sol
sashaice [31]

Answer:

800.0 mL.

Explanation:

  • To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.

<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>

M before dilution = 12.0 M, V before dilution = 100.0 mL.

M after dilution = 1.5 M, V after dilution = ??? mL.

∵ (MV)before dilution of HCl = (MV)after dilution of HCl

∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)

<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>

8 0
3 years ago
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