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trapecia [35]
2 years ago
6

What is the volume, in liters, occupied by 1.73 moles of N2 gas at 0.992 atm pressure and a temperature of 75º C? (R value- 0.08

206)
Chemistry
1 answer:
taurus [48]2 years ago
7 0

Volume of the nitrogen gas = 49.8 L

<u>Explanation:</u>

It is given that the pressure, number of moles and temperature of nitrogen gas, and gas constant value being constant and it is taken as 0.08206 L atm mol⁻¹K⁻¹.

Temperature = T = 75°C = 75 + 273 = 348 K

Pressure = P = 0.992 atm

Number of moles = n = 1.73 moles

We have to use the ideal gas equation, PV = nRT, and rearranging the equation to get Volume in litres.

V = $\frac{nRT}{P}

 = $\frac{1.73\times 0.08206\times348}{0.992}

= 49.8 L

So the volume of Nitrogen gas = 49.8 L

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1. What is the oxidation number for the silver ion in tarnish?
spayn [35]
Tarnish is Ag2S-silver sulfide and the oxidation state of silver is +1
7 0
2 years ago
How many miles of Fe2+ ions and MnO4- ions were titrated in each part 1 trial
Fudgin [204]

Answer:

In 1000 ml there is 0.10 moles of Fe 2+

Therefore, in 10 ml there is (0.1/1000)*10= 0.001 mol of Fe2+

mole ratio for rxn Fe2+ : MnO4- is

1 : 2

therefore if 0.001 moles of Fe2+ react then 0.001*2 =0.002 moles of MnO4- react with Fe2+

hence, molarity of MnO4- = (mol*vol)/1000

= 0.002*10.75/1000= 2.15*10-5M

Explanation:

Hope this helps

5 0
2 years ago
How many moles of nitrogen are in 3.7 moles of C8H11NO2?
Phoenix [80]
<h3>Answer:</h3>

               3.7 Moles of Nitrogen

<h3>Explanation:</h3>

                      On observing the chemical formula C₈H₁₁NO₂ (might be formula of Dopamine) it is found that one mole of this compound contains;

8 Moles of Carbon

11 Moles of hydrogen

1 Mole of Nitrogen and

2 Moles of Oxygen respectively.

<u>Calculate Number of Moles of Nitrogen:</u>

As,

                   1 Mole of C₈H₁₁NO₂ contains  =  1 Mole of Nitrogen

So,

            3.7 Moles of C₈H₁₁NO₂ will contain  =  X Moles of Nitrogen

Solving for X,

                       X  =  (3.7 Moles × 1 Mole) ÷ 1 Mole

                       X  =  3.7 Moles of Nitrogen

4 0
2 years ago
Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equil
Illusion [34]

Answer:

Part A

K = (K₂)²

K = (K₃)⁻²

Part B

K = √(Ka/Kb)

Explanation:

Part A

The parent reaction is

2Al(s) + 3Br₂(l) ⇌ 2AlBr₃(s)

The equilibrium constant is given as

K = [AlBr₃]²/[Al]²[Br₂]³

2) Al(s) + (3/2) Br₂(l) ⇌ AlBr₃(s)

K₂ = [AlBr₃]/[Al][Br₂]¹•⁵

It is evident that

K = (K₂)²

3) AlBr₃(s) ⇌ Al(s) + 3/2 Br₂(l)

K₃ = [Al][Br₂]¹•⁵/[AlBr₃]

K = (K₃)⁻²

Part B

Parent reaction

S(s) + O₂(g) ⇌ SO₂(g)

K = [SO₂]/[S][O₂]

a) 2S(s) + 3O₂(g) ⇌ 2SO₃(g)

Ka = [SO₃]²/[S]²[O₂]³

[SO₃]² = Ka × [S]²[O₂]³

b) 2SO₂(g) + O₂(g) ⇌ 2 SO₃(g)

Kb = [SO₃]²/[SO₂]²[O₂]

[SO₃]² = Kb × [SO₂]²[O₂]

[SO₃]² = [SO₃]²

Hence,

Ka × [S]²[O₂]³ = Kb × [SO₂]²[O₂]

(Ka/Kb) = [SO₂]²[O₂]/[S]²[O₂]³

(Ka/Kb) = [SO₂]²/[S]²[O₂]²

(Ka/Kb) = {[SO₂]/[S][O₂]}²

Recall

K = [SO₂]/[S][O₂]

Hence,

(Ka/Kb) = K²

K = √(Ka/Kb)

Hope this Helps!!!

6 0
2 years ago
What is the molarity of a solution that contains 122g of MgSO4 n 3.5L of solution?​
tatiyna

Answer:

0.29mol/L or 0.29moldm⁻³

Explanation:

Given parameters:

Mass of MgSO₄ = 122g

Volume of solution = 3.5L

Molarity is simply the concentration of substances in a solution.

Molarity = number of moles/ Volume

>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.

Number of moles = mass/ molar mass

Molar mass of MgSO₄:

Atomic masses: Mg = 24g

S = 32g

O = 16g

Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol

= (24 + 32 + 64)g/mol

= 120g/mol

Number of moles = 122/120 = 1.02mol

>>>> From the given number of moles we can evaluate the Molarity using this equation:

Molarity = number of moles/ Volume

Molarity of MgSO₄ = 1.02mol/3.5L

= 0.29mol/L

IL = 1dm³

The Molarity of MgSO₄ = 0.29moldm⁻³

6 0
2 years ago
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