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2 years ago

Calculate the oxidation number of s in S2O8^2-

Chemistry

1 answer:

mixer [17]2 years ago

3 0

Given problem:

S₂O₈²⁻

Find the oxidation number of S;

Oxidation number presents the extent of oxidation of each atom of elements a molecular formular or formula unit or an ionic radical.

For radicals:

"the algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion "

S₂O₈²⁻; oxidation number of O is usually -2

2(S) + 8(-2) = -2

2S - 16 = -2

2S = -2 + 16

2S = 14

S = +7

The oxidation state of S in the radical is +7

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Large sharks eat many other marine life animals. The sharks and the animals they eat are all part of which level of organization

Sladkaya [172]

Answer:

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2 years ago

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How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?

lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

N₂ + 3 H₂ ------—> 2 NH₃

1 mol 3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

N₂ + 3 H₂ ------—> 2 NH₃

1 mol (28 g/mol) 3 mol(2g/mol)

28 g 6 g

28 grams of N₂ react with 6 g of H₂

if 28 grams of N₂ produces 6 g of H₂ so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

28 g of N₂ ≅ 6 g of H₂

X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

X g of N₂ = 28 g x 14.2 g / 6 g

X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

N₂ + 3 H₂ ------—> 2 NH₃

6 g 2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

6 g of H₂ ≅ 2 mol of NH₃

14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

X mol of NH₃= 14.2 g x 2 mol / 6 g

X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of NH₃

Molar mass of NH₃ = 14 + 3(1)

Molar mass of NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

mass in grams = 4.73 mole x 17 g/mol

mass in grams = 80.41 g

mass of NH₃= 80.41 g

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2 years ago

Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure?

balu736 [363]

Answer:

C 8.09 SO2 gas

Explanation:

As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:

PV = nRT

PV / RT = n

1atm*3L / 0.082atmL/molK*298K =¨

0.123 moles of gas you have.

Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).

Mass of 0.123 moles of these gases is:

O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong

N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong

SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible

CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong

Right answer is:

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2 years ago

If you were to assume a 95% yield for the formation of the phosphonium ion, how many milligrams of benzyl chloride and triphenyl

luda_lava [24]

I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.

First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,

For Benzyl chloride,
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{126580 g of benzyl chloride}{X}$

Solving for X,
X = $\frac{220 mg . 126580 mg}{353420 mg}$
X = 78.79 mg

For PPh₃:
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{262290 g of PPh3}{X}$

Solving for X,
X = $\frac{220 mg . 262290 mg}{353420 mg}$
X = 163.27 mg

Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,

when $\frac{95 percent}{100 percent}$ = $\frac{220 g}{X}$
Solving for X,

X = $\frac{220 . 100}{95}$ = 231.57 mg

Now, calculate reactants mass with respect to 231.57 mg
when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{78.79 g of benzyl chloride}{X}$
Solving for ,

X = $\frac{231.57 . 78.79}{220}$ = 82.93 mg of Benzyl chloride

when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{163.27 g of PPh3}{X}$
Solving for ,

X = $\frac{231.57 . 163.27}{220}$ = 171.85 mg of PPh3

So, reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.

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2 years ago

A gas in a rigid container at 25°C has a pressure of 0.96 atm. A change in temperature causes the pressure to increase to 1.25 a

m_a_m_a [10]

Gay-Lussacs law states that pressure of a gas is directly proportional to temperature when the volume is kept constant

P / T = k

where P - pressure , T - temperature in kelvin and k - constant

$\frac{P1}{T1} = \frac{P2}{T2}$

where parameters for the first instance are on the left side of the equation and parameters for the second instance are on the right side of the equation

T1 - 25 °C + 273 = 298 K

substituting the values in the equation

$\frac{0.96 atm}{298 K}= \frac{1.25 atm}{T2}$

T2 = 388 K

temperature in celcius - 388 K - 273 = 115 °C

answer is C. 115 °C

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2 years ago

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Calculate the oxidation number of s in S2O8^2-​

Calculate the oxidation number of s in S2O8^2-