How many liters is 3.3 moles of a gas?

Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "

Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

As Mass = density × volume

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

PV = nRt

V = nRT/P

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0

3 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

3 years ago

What is the difference between concentrated acid and a weak one?

makkiz [27]

Answer:

Difference between concentrated acid and weak acid :---

According to Arrhenius's theory the substances which easily get dissociated into H+ ions when dissolved in water are acids.

And the substance which readily gives H+ ions on dissociation (when dissolved in water) are Strong Acid. Examples are HCl , H2SO4 etc.

While,

Concentration of acid is just the value of pH. pH is the measurement of concentration of acid or base. The lower the pH, higher the concentration.

So strong acid is strong because it gives H+ ions readily in water and Concentration is the value of pH.

7 0

2 years ago

What's it called when an electron has absorbed alot of energy?

emmainna [20.7K]

Answer:

Excited state

Explanation:

Excited state is the name of the state when electrons absorb energy and move to a higher energy level.

Hope this helps!

8 0

3 years ago

What is the mass of oxygen that can be produced from 2.79 moles of lead(ll) nitrate

denis23 [38]

1.38 moles of oxygen

Explanation:

Thermal decomposition of Lead (II) nitrate is shown by the balanced equation below;

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

The mole ration of Lead (II) nitrate to oxygen is 2: 1

Therefore 2.76 moles of Lead (II) nitrate will lead to production of? moles of oxygen;

2: 1

2.76: x

Cross-multiply;

2x = 2.76 * 1

x = 2.76 / 2

x = 1.38

8 0

2 years ago