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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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3 years ago

6. Cuando se oxidan en el aire 12,120 g de vapor de Zinc se obtienen 15,084 g del óxido. ¿Cuál es la fórmula empírica del óxido?

Eddi Din [679]

Answer:

12120 g + O2 = 15084 g

m Zn = 12.120 Kg

m óxido = 15.084 Kg

1. calcular la masa de cinc en gramos

g = 12,120 Kg x 1000 = 12120 g de cinc

g = 15.084 Kg x 100 = 15084 g de oxígeno

2. calcular gramos de Oxigeno

g O = 15084 g - 12120 g = 2964 g O2

3. calcular % de Zn y O

%m/m ( m soluto / m solc.) x 100

%m/m (Zn) = ( 1210 g / 15084 g ) x 100

% m/m (Zn) = 80.35 % = 80.35 g

%m/m (O) = ( 2964 g / 15084 g ) x 100

% m/m (Zn) = 19.65 % = 19.65 g

4. Calcular moles de cada elemento

Zn: 80.35 g / 65.38 g/mol = 1.228 mol

O: 19.65 g / 16 g/mol = 1.228 mol

5. dividir entre el menor de los elementos

Zn: 1.228 mol / 1.228 mol = 1

O: 1.228 mol / 1.228 mol = 1

6. Fórmula empírica: ZnO

3 0

3 years ago

The formula for the nitrate ion: NO3 NO2 NO3+

tekilochka [14]

Answer:

NO3-

Explanation:

edge2020

3 0

3 years ago

Read 2 more answers

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

Consider the electrolysis of molten barium chloride (bacl2). write the half-reactions. include the states of each species.

aksik [14]

Molten barium chloride is separeted into two species :
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

<span>The anode is positive and the cathode is negative.</span>

7 0

3 years ago