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Snezhnost [94]
3 years ago
10

Balance it by oxidation number method:Zn +HNO3----Zn(NO3)2+NO+H2O​

Chemistry
2 answers:
Westkost [7]3 years ago
6 0

Answer:

Your ans is in the picture.

timofeeve [1]3 years ago
3 0

Answer:

Answer:

step 1:balance skeleton equation the chemical equation:

Zn +HNO3➔Zn(NO3)2+NO+H2O

step 2: identity undergoing oxidation or reduction

here

Zn➔Zn(NO3)2

Zn is oxidized from 0 to 2 in oxidation no.

HNO3➔NO

N is reduced from 5 to 2 in oxidation no

Step 3: calculate change in oxidation no.

change in oxidation no

in Zn=0-2=-2=2

in

N=5-2=3

Step 4: Balance it by doing crisscrossed multiplication

we get;

3Zn +2HNO3➔3Zn(NO3)2+2NO+H2O

step 6:Balance other atoms except H & O

3Zn +2HNO3➔3Zn(NO3)2+2NO+H2O

3Zn +2HNO3+6HNO3➔3Zn(NO3)2+2NO+H2O

finally: balance H

<em><u>3Zn +8HNO3➔3Zn(NO3)2+2NO+4H2O</u></em>

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In which of the titrations described below will the first (or only) equivalence point be reached upon the addition of 25.0 mL of
Mnenie [13.5K]

Answer:

1, 3, and 4

Step-by-step explanation:

We must calculate the volume of NaOH needed for each titration.

<em>1) HCl </em>

HCl + NaOH ⟶ NaCl + H₂O

n(HCl)       = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH)  = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

<em>2) H₂C₂O₄ </em>

H₂C₂O₄ + NaOH ⟶ NaHC₂O₄

n(H₂C₂O₄) = 50.0 mL × (0.100 mmol/1mL) = 5.00 mmol

n(NaOH)    = 5.00 mmol H₂C₂O₄ × (1 mmol NaOH/1 mmol H₂C₂O₄)

= 5.00 mmol NaOH

V(NaOH)   = 5.00 mmol × (1 mL/0.100 mmol) = 50.0 mL

<em>3) HC₂H₃O₂ </em>

HC₂H₃O₂ + NaOH ⟶ NaC₂H₃O₂ + H₂O

n(HC₂H₃O₂) = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

<em>4) HBr </em>

HBr + NaOH ⟶ NaBr + H₂O

n(HBr) = 25.0 mL × (0.100 mmol/1mL) =2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

Titrations 1, 3, and 4 reach the first or only equivalence point  at 25.0 mL NaOH.

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