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3 years ago

According to Charles's law, under which conditions will the volume of a given amount of gas be proportional to temperature?

2 answers:

7 0

The Charles's Law states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant. Hope this helps. Have a wonderful day.

Pavel [41]3 years ago

5 0

When the pressure is kept constant

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A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture

Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

$n=\frac{PV}{RT}$

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

$n=(\frac{1.45*2.50}{0.0821*293})$

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = $8.60g(\frac{1mol}{44g})$

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

$X=\frac{0.258}{2}$

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane = $\frac{moles of Methane}{total moles}$

mole fraction of methane = $\frac{0.129}{0.151}$

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

6 0

3 years ago

I need help with this question

denis-greek [22]

It’s B. I had that to

6 0

4 years ago

when 6g acetic acid is dissolved in 1000cm3 of solution then how many molecules ionize out of 1000 acetic acid molecules

iVinArrow [24]

Answer:

24.8 molecules are ionized from 1000 acetic acid molecules.

Explanation:

Acetic acid, CH₃COOH dissociates in water, thus:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Ka = 6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]

<em>That means amount of CH₃COO⁻ (the dissociated form) that are produced is followed by the equilibrium of the weak acid.</em>

<em />

The initial molar concentration of acetic acid (Molar mass: 60g/mol) is:

6g ₓ (1mol / 60g) = 0.1 moles acetic acid, in 1000cm³ = 1L.

0.1 moles / L = <em>0.1M</em>

The 0.1M of acetic acid will dissociate producing X of CH₃COO⁻ and H⁺, thus:

[CH₃COOH] = 0.1M - X

[CH₃COO⁻] = X

[H⁺] = X

Replacing in Ka formula:

6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]

6.3x10⁻⁵ = [X] [X] / [0.1 - X]

6.3x10⁻⁶ - 6.3x10⁻⁵X = X²

6.3x10⁻⁶ - 6.3x10⁻⁵X - X² = 0

Solving for X

X = - 0.0025 → False solution, there is no negative concentrations.

X = 0.00248M

That means, a 0.1M of acetic acid produce:

[CH₃COO⁻] = X = 0.00248M solution of the ionized form.

In a basis of 1000 molecules:

1000 molecules × (0.00248M / 0.1M) = 24.8

<h3>24.8 molecules are ionized from 1000 acetic acid molecules.</h3>

8 0

3 years ago

If a solution containing 117.63 g of silver chlorate is allowed to react completely with a solution containing 10.23 g of lithiu

GuDViN [60]

Answer:

18.8 g

Explanation:

The equation of the reaction is;

AgClO3(aq) + LiBr(aq)------>LiClO3(aq) + AgBr(s)

Number of moles of AgClO3 = 117.63 g/191.32 g/mol = 0.6 moles

Number of moles of LiBr = 10.23 g/86.845 g/mol = 0.1 moles

Since the molar ratio is 1:1, LiBr is the limiting reactant

Molar mass of solid AgBr = 187.77 g/mol

Mass of precipitate formed = 0.1 moles * 187.77 g/mol

Mass of precipitate formed = 18.8 g

8 0

3 years ago

Number 51 is 0.00150 ml to l

xenn [34]

Answer:

The answer to your question is

51.- 1.59 x 10⁻⁶

53.- NaHCO₃ + HCl ⇒ H₂O + CO₂ + NaCl

Explanation:

51.- Convert 0.00159 ml to liters

Use a rule of three

1000 ml -------------- 1 l

0.00159 ml --------- x

x = (0.00159 x 1) / 1000

x = 0.00000159 l = 1.59 x 10⁻⁶ l

53.- Sodium hydrogen carbonate (NaHCO₃) reacts with hydrochloric acid to produce salt, water and carbon dioxide.

Sodium hydrogen carbonate = NaHCO₃

Hydrochloric acid = HCl

Reaction

NaHCO₃ + HCl ⇒ H₂O + CO₂ + NaCl

water = H₂O

carbon dioxide = CO₂

salt = NaCl

8 0

3 years ago