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3 years ago

30 its ASAP!!!! What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?

1 answer:

5 0

Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)

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2 years ago

When solid ammonium chloride dissociates at a certain temperature in a 0.500 dm3 container, ammonia and hydrogen chloride are fo

SIZIF [17.4K]

1) Reaction

<span>NH4Cl(s) ---> NH3(g) + HCl(g)

2) equilibrium equation, Kc

Kc = [NH3] * [HCl]

3) Table of equilibrium formation

step    concentrations
   </span>
<span>      NH4Cl(s)    NH3(g) HCl(g)

start 1.000 mole    0 0

react - x

produce +x    + x
   ------------------ ---------- -----------

end     1 - x     +x    +x


1 - x = 0.3 => x = 1 - 0.3 = 0.7

[NH3] = [HCl] = 0.7/0.5 liter = 1.4 (I used 0.500 dm^3 = 0.5 liter)

4) Equilibrium equation:

Kc = [NH3] [HCl] = (1.4)^2 = 1.96

Which is the number that you were looking for.

Answer: Kc = 1.96
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8 0

3 years ago

Can someone plz help me im running out of time !!!!

ankoles [38]

Answer:

1. C(2)+H2(1) -> C2H6(1)

2. NH3(2)+O2(3)-> HCN(2)+H2O(3)

I am not sure about the second one.

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2 years ago

If you need 2.5 moles of Rb2Cr7, how many grams do you need?

BartSMP [9]

Answer:

Explanation:

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2 years ago

In the unbalanced chemical reaction for the combustion of propane determine at standard temperature and pressure how many liters

Semmy [17]

Answer:

9 liters of CO₂ are produced by this combustion

Explanation:

In order to determine the volume of produced CO₂, we start with the reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

We need, O₂ density to find out the mass, that has reacted.

δ O₂ = O₂ mass / O₂ volume → δ O₂ . O₂ volume = O₂ mass

δ O₂ = 1.429 g /dm₃ (1dm³ = 1L) 1.429 g/L . 15L = 21.4 g of O₂

We convert the mass to moles: 21.4 g . 1mol / 32 g = 0.670 moles

By stoichiometry, 5 moles of O₂ can produce 3 moles of CO₂

Then, 0.670 moles of O₂ will produce (0.670 . 3) /5 = 0.402 moles of dioxide.

We apply Ideal Gases Law for STP, to find out the CO₂ volume

V = (n . R . T) / P → V = (0.402 mol . 0.082 . 273K) / 1 atm = 8.99 L ≅ 9 L

4 0

3 years ago