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Roman55 [17]
3 years ago
12

Select all that apply. At STP conditions, 0.50 moles of CO(g), Cl2(g), O2(g): will contain the same number of molecules, will co

ntain the same number of atoms, will occupy the same volume, will have the same g.f.w.
Chemistry
1 answer:
Mariana [72]3 years ago
6 0
1 mol is (always) the same number of units: 6.02 * 10^23 units.


So 0.5 moles  of any gas has the same number of molecules.


Also, we know by the ideal gas laws that a give number of molecules of any gas will occupy the same volume.


Given that the three gases have the same number of atoms in the molecular fomula (2), 0.5 moles will also have the same number of atoms.


g.f.w. stands for grams formula weight and that is different for all of them, because gfw is calcualted from the atomics masses of each atom in the molecule.


Then at STP (standard temperature and pressure) conditions 0.50 moles of any of the gas:


- will contain the same number of molecules,

- will contain the same number of atoms

- will occupy the same volume.

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Ammonium hydrogen sulfide NH4HS(s) decomposes on heating according to the reaction NH4HS(s) ↔ NH3(g) + H2S(g) At 25 °C the equil
yan [13]

Answer:

0.66atm

Explanation:

Based on the reaction:

NH₄HS(s) ↔ NH₃(g) + H₂S(g)

The equilibrium constant, Kp, is defined as:

Kp = 0.11 = P_{NH_3} P_{H_2S}

As moles of gas produced for NH₃(g) and H₂S(g) are the same, it is possible to write:

0.11 = P_{NH_3}^2

0.33 = P_{NH_3}

That means pressure of NH₃(g) is 0.33atm and H₂S(g) is, also, 0.33atm. Thus, total pressure is:

0.33atm×2 = <em>0.66atm</em>

3 0
4 years ago
Read 2 more answers
El acido nitrico se obtiene industrialmente por oxidacion de amoniaco (nh3), de acuerdo con la ecuacion: 4nh3 + 702 - h2o + hno2
devlian [24]

Answer:

2800 g de ácido nítrico

Explanation:

La ecuación por la oxidación de amoniaco es:

4NH₃  +  7O₂  →  4H₂O  +  2HNO₂  +  2HNO₃

Si pensamos que el oxígeno es el reactivo limitante, trabajamos con el amoniaco. Convertimos su masa a moles:

1.36 kg = 1360 g

1360 g . 1mol  /17g = 80 moles

Si 4 moles de amoniaco pueden producir 2 moles de acido nítrico

80 moles producirán, (80 . 2)/4 = 40 moles.

Convertimos los moles a gramos:

40 mol . 63g /mol = 2520 g

Si le aplicamos la pureza

2520 g . 100/90 = 2800 g

6 0
3 years ago
It is necessary to make 225 mL of 0.222 M solution of nitric acid. Looking on the shelf, you see only 16 M nitric acid. How much
nalin [4]

Explanation:

The required concentration of HNO_3 M1 =0.222 M.

The required volume of HNO_3 is V1 =225 mL.

The standard solution of HNO_3 is M2 =16 M.

The volume of standard solution required can be calculated as shown below:

Since the number of moles of solute does not change on dilution.

The number of moles n=molarity *  volume

M_1.V_1=M_2.V_2

V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL

Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.

6 0
3 years ago
How many moles of potassium are in 78.20 g?
Fynjy0 [20]
The answer to this question would be: 2 mol

To answer this question, you need to know the molecular weight of Potassium. Molecular weight determines how much the weight of 1 mol of a molecule has. 
Potassium or Kalium molecular weight is 39.1 gram/mol. Then, 78.20gram of potassium should be: 78.20g/ (39.1g/mol)= 2 mol
7 0
3 years ago
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How do you dilute a concentrated solution?
AVprozaik [17]

Answer:

add more water

Explanation:

5 0
3 years ago
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