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Burka [1]
3 years ago
14

What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy

Chemistry
1 answer:
ivanzaharov [21]3 years ago
4 0
Data Given:
Moles = n = 3.91 mol

Pressure = P = 5.35 atm

Temperature = T = 323 K

Volume = V = ?

Formula used: Ideal Gas Equation is used,

P V = n R T
Solving for V,
V = n R T / P
Putting Values,
V = (3.91 mol × 0.0825 atm.L.mol⁻¹.K⁻¹ × 323 K) ÷ 5.35 atm

V = 19.36 L
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A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. Wh
RSB [31]

Answer:

C2H5NO

Explanation:

constituent elements                    N              O               C                H

Mass composition                     0.420        0.480         0.540         0.135

mole ratio                                   0.42/14       0.48/16      0.54/12         0.135/1

                                              = 0.03                0.03             0.045         0.135

dividing by the smallest           0.03/0.03     0.03/0.03    0.045/0.03 0.135/0.03

ratio                                      =        1                      1                1.5                 4.5

                                             =        1                      1                  2                   5

EMPERICAL FORMULA = C2H5NO

7 0
3 years ago
Determine the empirical formula of a
k0ka [10]

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold  = 2.6444g

Mass of Chlorine  = 0.476g

Unknown:

Empirical formula  = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements                                     Au                                             Cl

Mass                                         2.6444                                     0.476

Molar mass                                 197                                          35.5

Number of moles                  2.6444/197                                 0.476/35.5

                                                 0.013                                           0.013

Divide by the

smallest                                 0.013/0.013                                 0.013/0.013

                                                       1                                                   1

The empirical formula of the compound is AuCl

3 0
2 years ago
Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo
Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>

4 0
3 years ago
Should you place materials that have toxic fumes by an open window?
alexandr402 [8]
No. You shouldn't place materials that have toxic fumes because it can get in people's lunges through breathing.
3 0
3 years ago
Please answer the chemistry question shown in the attachment photo! 25 points!
iogann1982 [59]
I’m so sorry if it’s wrong but I think it’s this.....

Answer:

2NaHCO3(s) = Na2CO3(s) + CO2(g) + H2O(g).

And I found it on Quora.com
By the way NaHCO3 is Sodium bicarbonate but a.k.a Baking soda
Hoped this helped :)

5 0
3 years ago
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