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ser-zykov [4K]
2 years ago
7

What is the oxidation state of selenium in SeO3?​

Chemistry
1 answer:
ra1l [238]2 years ago
8 0

Answer: The oxidation state of selenium in SeO3 is +6

Explanation:

SeO3 is the chemical formula for selenium trioxide.

- The oxidation state of SeO3 = 0 (since it is stable and with no charge)

- the oxidation number of oxygen (O) IN SeO3 is -2

- the oxidation state of selenium in SeO3 = Z (let unknown value be Z)

Hence, SeO3 = 0

Z + (-2 x 3) = 0

Z + (-6) = 0

Z - 6 = 0

Z = 0 + 6

Z = +6

Thus, the oxidation state of selenium in SeO3 is +6

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A certain gas is present in a 12.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas
diamong [38]
There are 3 parts in this question:
1) To find the initial Boyle's constant k_{i}
2) To find the final Boyle's constant k_{f}
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = V_{i} = 12.0 L
The initial pressure(in atmospheric pressure) = P_{i} = 4.0 atm

The final pressure(in atmospheric pressure) = P_{f} = 8.0 atm
The final volume of the cylinder(in litres) = V_{f} = 6.0 L

First you need to know what Boyle's law is:
<span>Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
</span>
The Mathematical form of Boyle's law is:
P =  \frac{k}{V}

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1. 
The initial volume of the cylinder(in litres) = V_{i} = 12.0 L
The initial pressure(in atmospheric pressure) = P_{i} = 4.0 atm
The Boyle's constant = k_{i} = ?

According to the Boyle's law,

P_{i} = \frac{k_{i}}{V_{i}}

=> k_{i} =  P_{i}V_{i}
Plug-in the values in the above equation, you would get:
k_{i} = 4.0 * 12.0 = 48

Ans-1) k_{i} = 48

2.
The final pressure(in atmospheric pressure) = P_{f} = 8.0 atm
The final volume of the cylinder(in litres) = V_{f} = 6.0 L
The Boyle's constant = k_{f} = ?

According to the Boyle's law,

P_{f} = \frac{k_{f}}{V_{f}}

=> k_{f} =  P_{f}V_{f}
Plug-in the values in the above equation, you would get:
k_{f} = 8.0 * 6.0 = 48

Ans-2) k_{f} = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

k_{i} = k_{f}

Since,
k_{i} = 48
k_{f} = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

-i

4 0
3 years ago
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Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to
valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

<u>Explanation</u>:

  • Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

                        Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

  • From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of  H₂SO₄.
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5 0
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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t
Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

\Delta H_{f,CuO}=-157.3 kJ/mol

\Delta H_{f,NO_2}= 33.2 kJ/mol

\Delta H_{f,O_2}= 0 kJ/mol (standard state)

\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]

The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)

\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

6 0
2 years ago
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Idk there just making me answer it
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2 years ago
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2 years ago
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