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schepotkina [342]
1 year ago
10

Question 1 how many moles of helium are found in a balloon that contains 5.5 l of helium at a pressure of 1.15 atm and a tempera

ture of 22.0 ºc?
Chemistry
2 answers:
DaniilM [7]1 year ago
7 0

Answer : The number of moles of helium gas found in a balloon are 0.26 moles.

Explanation :

To calculate the moles of helium gas, we use the equation given by ideal gas :

PV = nRT

where,

P = Pressure of helium gas = 1.15 atm

V = Volume of the helium gas = 5.5 L

n = number of moles of helium gas = ?

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

T = Temperature of helium gas = 22.0^oC=273+22.0=295K

Putting values in above equation, we get:

1.15atm\times 5.5L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 295K\\\\n=0.26mol

Therefore, the number of moles of helium gas found in a balloon are 0.26 moles.

denpristay [2]1 year ago
3 0
Using pv=nRT
T= 22+273 K
P=1.15 Pa
R=8.31
V =5.5... (u did not write the unit so...)

Therefore n,mole= (1.15×5.5) ÷ (273+22)(8.31)
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The molecular mass of methyl ethanoate is 74.1 amu . calculate the molecular mass of propanoic acid, an isomer of methyl ethanoa
kobusy [5.1K]
Isomers are the compounds having same molecular formula but different structural formula. 

Since, molecular formula is isomers are same, they have same mass.

Now, <span>methyl ethanoate is  an isomer of propanoic acid, hence they have same mass.

</span>∴ Molecular mass of propanoic acid = 74.1 amu or 74.1 g/mol<span>
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4 0
2 years ago
Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57
vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

<em />

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0
1 year ago
When aqueous solutions of NH4OH(aq) and CuCl2(aq) are mixed, the products are NH4Cl(aq) and Cu(OH)2(s). What is the net ionic eq
Ilia_Sergeevich [38]

Answer:

2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)

Explanation:

The net ionic equation usually shows the main ionic reaction that goes in the system. The other ions that do not participate in this net ionic equation are called spectator ions. Spectator ions do not participate in the main reaction occurring in the system.

The net ionic equation quite often result in the formation of a solid precipitate in the system such as Cu(OH)2.

The net ionic equation for this reaction is;

2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)

5 0
2 years ago
A solution is prepared by adding 100 mL of 1.0 M HC2H3O2 (aq) to 100 mL of 1.0 M NaC2H3O2 (aq). The solution is stirred and its
DIA [1.3K]

Answer:

(C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l)

Explanation:

A buffer is a solution of a weak acid and its salt. It mitigates against changes in acidity or alkalinity of a system. A buffer maintains the pH at a constant value by switching the equilibrium concentration of the conjugate acid or conjugate base respectively.

Addition if an acid shifts the equilibrium position towards the conjugate acid side while addition of a base shifts the equilibrium position towards the conjugate base side.

5 0
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Answer:

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Explanation:

Charge on 1 electron = -1.6\times 10^{-19}\ C

The expression for charge is:-

Charge=n\times q_e

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-4.8\times 10^{-6}=n\times (-1.6\times 10^{-19})

n=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}

Total number of electrons, n = 3.0\times 10^{13}

5 0
1 year ago
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