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aniked [119]
3 years ago
15

Calculate the volume in L of 11.6 moles of Neon at 120 K when it has a pressure of 25.9 atm

Chemistry
1 answer:
beks73 [17]3 years ago
7 0

Answer:

The volume of the gas is approximately 4.41 liters

Explanation:

The details of the data of the Neon gas are;

The number of moles of Neon gas present, n = 11.6 moles

The temperature of the sample of Neon gas, T = 120 K

The pressure of the sample of the Neon gas, P = 25.6 atm

By the ideal gas equation, we have;

P·V = n·R·T

Where;

R = The universal gal constant = 0.08205 L·atm·mol⁻¹·K⁻¹

Therefore, we get;

V = n·R·T/P

Which gives;

V = 11.6 moles × 0.08205 L·atm·mol⁻¹·K⁻¹ × 120 K/(25.9 atm) ≈ 4.4097915 L

The volume of the gas, V ≈ 4.41 L.

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Write a balanced molecular equation for hydrochloride acid solution combined with a potassium hydroxide solution, an acid base r
qaws [65]

Answer:

HCl (aq) + KOH (aq) --------> KCl (aq) + H2O (l)   [balanced molecular equation]

Explanation:

Balanced molecular equation has all the components of reaction written as "molecules" thus the name molecular (the other equations break appropriate components into ions, those are total ionic or net ionic equation)

We know hydrochloride acid solution is written as HCl (aq) and potassium hydroxide solution is written as KOH (aq). We know from our knowledge that when acids react with bases they make salt and water, so now we react!

HCl (aq) + KOH (aq) --------> KCl (aq) + H2O (l)   [balanced molecular equation]

7 0
3 years ago
Formula of preparation of sodium hydroxide​
Umnica [9.8K]

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It is an ionic compound consisting of sodium cation (Na+) and hydroxide (OH-) anion.

7 0
2 years ago
An enclosed vessel contains 2.5g of 9b nitrogen and 13.3g of chlorine at s.T.P. Of What will be the partial pressure of the Il n
kow [346]

Answer:

0.535 atm

Explanation:

Since the volume of the tank is constant, we use Gay- Lussac's law to find the pressure at 180°C.

So, P₁/T₁ = P₂/T₂ where P₁ = pressure at S.T.P = 1 atm, T₁ = temperature at S.T.P = 273.15 K, P₂ = pressure of gas at 180 °C and T₂ = 180 °C = 273.15 + 180 K = 453.15 K

So, P₁/T₁ = P₂/T₂

P₂ = P₁T₂/T₁

Substituting the values of the variables into the equation, we have

P₂ = P₁T₂/T₁

P₂ = 1 atm × 453.15 K/273.15 K

P₂ = 1 atm × 1.66

P₂ = 1.66 atm

We now need to find the total number of moles of each gas present

number of moles of nitrogen = mass of nitrogen, m/molar mass of nitrogen molecule M

n = m/M

m = 2.5 g and M = 2 × atomic mass of nitrogen (since it is diatomic) = 2 × 14 g/mol = 28 g/mol

So, n = 2.5 g/28 g/mol

n = 0.089 mol

number of moles of chlorine, n' = mass of chlorine, m'/molar mass of chlorine molecule M'

n' = m'/M'

m' = 13.3 g and M = 2 × atomic mass of chlorine (since it is diatomic) = 2 × 35.5 g/mol = 71 g/mol

So, n' = 13.3 g/71 g/mol

n' = 0.187 mol

So, the total number of moles of gas present is n" = n + n' = 0.089 mol + 0.187 mol = 0.276 mol

So, the partial pressure due to nitrogen gas, P = mole fraction of nitrogen × pressure of gas at 180 °C

P = n/n" × P₂

P = 0.089 mol/0.276 mol × 1.66 atm

P = 0.322 × 1.66 atm

P = 0.535 atm

8 0
3 years ago
What is the approximate percent by mass of oxygen in C2H4O2
kherson [118]

Answer:

Approximately 53.3  %.

Explanation:

Molar mass of C2H4O2

= 2 * 12.011 + 4 * 1.008 + 2 * 15.999

=  60.052

% oxygen = (31.998* 100) / 60.052 = 53.28 %

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