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3 years ago

Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?

2 answers:

5 0

The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

Crazy boy [7]3 years ago

3 0

Answer:

No element decreases its oxidation number.

Explanation:

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Hoochie [10]

Answer:

(a) 156.41 (b) 10 and 14 (c) 16

Explanation:

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2 years ago

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BARSIC [14]

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3 years ago

It the mass of a material is 46 grams and the volume of the material is 8 cm ^3 What would the density of the material to be

Nat2105 [25]

5.75 Grams per cm^3

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2 years ago

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = 240.9

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = 144

Now we can find percentage composition / percentage by mass of oxygen.

% composition = $\frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound}$ × 100

% composition = $\frac{144}{240.9}$ × 100 = 59.776%

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0

2 years ago

Two closed vessels contain chlorine gas at the same conditions of temperature and

wariber [46]

Answer:12 mol

Explanation: both vessels are at the same temp and pressure (and the pressure is low and/or the temperature high).

6.7mol per 1.3L = 6.7/1.3 mol/L

so in 2.33L = 6.7*2.33/1.3 = 12 mol

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2 years ago