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2 years ago

What is the oxidation state of each element in FeBr2?

1 answer:

5 0

The oxidation states of the elements Fe and Br needs to be determined.
Fe is a transition element therefore is capable of multiple oxidation states.
an ionic compound is made with Fe as the cation and halide ion as anion
Br⁻ which is a halogen has a charge of -1.
In ions the charge of the ion is equal to the oxidation state of the ion. Br⁻ has oxidation state of -1.
FeBr₂ is a compound therefore net charge is 0.
the sum of the products of the oxidation states of the individual elements by the number of atoms should add upto the net charge
oxidation number of Fe cation + (oxidation number of Br⁻ x 2) = 0
say the oxidation state of Fe is X and oxidation number of Br⁻ is -1
X + (2 x -1)= 0
X - 2 = 0
X = 2
this means that oxidation number of Fe is +2
and oxidation number of Br is -1

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Consider the following balanced redox reaction: 2CrO2-(aq) + 2H2O(l) + 6ClO-(aq) LaTeX: \longrightarrow⟶ 2CrO42-(aq) + 3Cl2(g) +

frutty [35]

Answer:

1. Chromium

2. Chlorine.

3. Chlorine.

4. Chromium.

5. 12 electrons.

Explanation:

Hello,

In this case, the given reaction with the appropriate oxidation states turns out:

$2(Cr^{+3}O^{-2}_2)^-(aq) + 2H_2O(l) + 6(Cl^{+1}O^{-2})^-(aq)\longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq) + 3Cl^0_2(g) + 4OH^-(aq)$

In such a way, the oxidation half-reaction is written for chromium as the reducing agent so it is oxidized from +3 to +6, nonetheless, since there are two chromiums undergoing such change, 6 electrons are being transferred as shown below:

$2(Cr^{+3}O^{-2}_2)^-(aq) \longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq)+6e^-$

On the other hand, chlorine's reduction half-reaction as the oxidizing agent result from the transfer of 6 electrons as well from +1 to 0, nonetheless, there are 6 chlorines undergoing such change:

$6(Cl^{+1}O^{-2})^-+6e^-\longrightarrow 3Cl^0_2(g)$

Therefore, there are 12 electrons that are being transferred, 6 for chromium and 6 for chlorine.

Best regards.

5 0

2 years ago

The interior of a refrigerator has a volume of 0.600 m3. the temperature inside the refrigerator in 282 k, and the pressure is 1

Dafna11 [192]

In this instance we can use the ideal gas law equation to find the number of moles of gas inside the refrigerator
PV = nRT
where
P - pressure - 101 000 Pa
V - volume - 0.600 m³
n - number of moles
R - universal gas constant - 8.314 J/mol.K
T - temperature - 282 K
substituting these values in the equation
101 000 Pa x 0.600 m³ = n x 8.314 J/mol.K x 282 K
n = 25.8 mol
there are 25.8 mol of the gas
to find the mass of gas
mass of gas = number of moles x molar mass of gas
mass = 25.8 mol x 29 g/mol = 748.2 g
mass of gas present is 748.2 g

6 0

2 years ago

What reaction is depicted by the given equation: Au3+ + 3e− Au

mr Goodwill [35]

Reduction. B is the correct answer

3 0

2 years ago

Read 2 more answers

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

2 years ago

A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solu

irinina [24]

Answer:

6,216.684 kilograms of sodium carbonate must be added to neutralize $6.05\times 10^3 kg$ of sulfuric acid solution.

Explanation:

Mass of sulfuric acid solution = $6.05\times 10^3 kg=6.05\times 10^6 g$

$1 kg = 10^3 g$

Percentage mass of sulfuric acid = 95.0%

Mass of sulfuric acid = $\frac{95.0}{100}\times 6.05\times 10^6 g$

$=5,747,500 g$

Moles of sulfuric acid = $\frac{5,747,500 g}{98 g/mol}=58,647.96 mol$

$H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O$

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.

Then 58,647.96 moles of sulfuric acisd will be neutralized by :

$\frac{1}{1}\times 58,647.96 mol=58,647.96 mol$ of sodium carbonate

Mass of 58,647.96 moles of sodium carbonate :

$106 g/mol\times 58,647.96 mol=6,216,683.76 g$

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg

6,216.684 kilograms of sodium carbonate must be added to neutralize $6.05\times 10^3 kg$ of sulfuric acid solution.

3 0

2 years ago