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4 years ago

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For each reaction, identify the element that gets reduced and the element that gets oxidized. 2AgCl+Zn⟶2Ag+ZnCl2 Identify the el

ement that gets reduced. chlorine zinc silver Identify the element that gets oxidized. chlorine silver zinc 4NH3+3O2⟶2N2+6H2O Identify the element that gets reduced. hydrogen oxygen nitrogen Identify the element that gets oxidized. nitrogen oxygen hydrogen Fe2O3+2Al⟶Al2O3+2Fe Identify the element that gets reduced. oxygen aluminum iron Identify the element that gets oxidized. aluminum oxygen iron

1 answer:

PolarNik [594]4 years ago

7 0

Answer:

Explanation:

Oxidation no is equal to charge on each atomic ion. If it is increased , element is oxidised and if it is decreased , element is reduced .

2AgCl+Zn⟶2Ag+ZnCl2

Zinc is oxidised , Ag is reduced .

Ag⁺ converts to Ag . ( oxidation number is reduced ) so Ag is reduced.

Zn converts to Zn⁺² ( oxidation number is increased ) so Zn is oxidised .

4NH₃+3O₂⟶2N₂+6H₂O

oxidation number of nitrogen in ammonia is - 3

oxidation no of nitrogen in nitrogen is zero.

Oxidation no of nitrogen is increased so it is oxidised.

oxidation no of oxygen is zero in oxygen and its oxidation no in water is -2 . So oxidation no is reduced so oxidation is reduced.

Fe₂O₃+2Al⟶Al₂O₃+2Fe

oxidation no of Fe in Fe₂O₃ is + 3 and it is zero in Fe so iron is reduced.

oxidation no of Al in Al is zero and it is +3 in Al₂O₃ so it is oxidised .

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sveticcg [70]

Answer:

23.92 g

Explanation:

Molar mass of H2SO4 = (2×1)+32+(16×4)= 2+32+48= 82g/mol

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

I mole of H2SO4 = 2 moles of NaOH

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3 years ago

How many milliliters of 2.19 M H2SO4 are required to react with 4.75 g of solid containing 21.6 wt% Ba(NO3)2 if the reaction is

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Answer:

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Explanation:

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This solid contains 21.6 wt% $Ba(NO_3)_2$ , thus,

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The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.026\ g}{261.337\ g/mol}$

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Considering the reaction as:

$Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3$

1 moles of $Ba(NO_3)_2$ react with 1 mole of $H_2SO_4$

Thus,

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Moles of $H_2SO_4$ = 0.003926 mole

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$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

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So,

$2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}$

Volume = 0.0017927 L

Also, 1 L = 1000 mL

<u>So, volume = 1.7927 mL</u>

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