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1 year ago

How many molecules of propane were in the erlenmeyer flask? Avogadro's number is 6. 022 × 10^23 molecules/mol

Chemistry

1 answer:

Ede4ka [16]1 year ago

4 0

3.74× $10^{21}$

3.74 × $10^{21}$ molecules of propane were in the erlenmeyer flask.

number of moles of propane can be calculated as moles of propane.

mass of propane = 0.274 g

molar mass of propane = 44.1

So this gives us the value of 6.21× $10^{-3}$ moles of propane

No one mole of propane As a 6.0-2 × $10^{23}$

so, 6.21 × $10^{-3}$ × 6. 022 × 10^23

= 3.74 × $10^{21}$

Therefore, molecules of propane were in the erlenmeyer flask is found to be 3.74 × $10^{21}$

<h3>What is erlenmeyer flask?</h3>

A laboratory flask with a flat bottom, a conical body, and a cylindrical neck is known as an Erlenmeyer flask, sometimes known as a conical flask or a titration flask.
It bears the name Emil Erlenmeyer after the German chemist.

<h3>What purpose does an Erlenmeyer flask serve?</h3>

Liquids are contained in Erlenmeyer flasks, which are also used for mixing, heating, chilling, incubating, filtering, storing, and other liquid-handling procedures.

For titrations and boiling liquids, their sloped sides and small necks make it possible to whirl the contents without worrying about spills.

To learn more about calculating total molecules visit:

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4.One rubber tire can generate about250,000 BTUS (1BTU=.0003KW ) when it is burned. The average American home consumes about 10,

jekas [21]

Answer:

The number of tires required to power ten homes for one year is approximately 2,730 tires

Explanation:

The given information are;

The energy generated by burning one rubber tire = 250,000 BTUs

1 BTU = 0.0003 kW

The amount of energy consumed by the average American home = 10,000 kWh

Therefore, the amount of energy generated by burning one tire in kW can be found as follows;

The energy generated by burning one rubber tire = 250,000 BTUs

1 BTU = 0.0003 kW

∴ 1 BTU × 250,000 = 250,000 BTUs = 0.0003 kW × 250,000 = 75 kW

We note that 250,000 BTUs is equivalent to 263764 kJ

1 kWh is equivalent to 3600 kJ

10,000 kWh = 10,000 × 3600 = 36,000,000 kJ

The energy requirement for ten homes for one year = 10 × 36,000,000 kJ = 360,000,000 kJ

At 50% efficiency, the energy produced per tire = 0.5 × 263764 kJ = 131,882 kJ

The electrical energy produced per tire = 131,882 kJ/tire

The number of tires required to power ten homes for one year = 360,000,000 kJ/ 131,882 kJ/tire = 2,729.71293 ≈2,730 tires

The number of tires required to power ten homes ≈2,730 tires.

8 0

3 years ago

A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting

n200080 [17]

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

= 0.15 x 0.3/0.8

= 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

= 0.10 x 0.5/0.8

= 0.0625 M

Note: Total volume = 500 + 300 = 800 mL = 0.8 dm3

pH = 3.14266 + log [0.05625/0.0625]

= 3.14267 + (-0.04575749056)

= 3.09691250944

From all the available options below:

a) 2.97

b) 3.10

c) 3.19

d) 3.22

e) 3.32

The correct option is b.

4 0

3 years ago

a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel

eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

1) What is the partial pressure of SO₂ gas in the larger container?

∵ P = nRT/V.

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

2) What is the partial pressure of N₂ gas in the larger container?

∵ P = nRT/V.

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

3) What is the total pressure in the vessel?

According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0

3 years ago

An airplane starts at rest and Excelerator down the runway for 20 seconds. At the end of the runway, its velocity is 80 m/s Nort

masha68 [24]

According to the formula you have given us to work with . . .

1). The airplane's acceleration is

(80 m/s north - zero) / (20 sec) = 4 m/sec^2 north

2). For the cyclist:

(V-final - zero) / 20sec = 0.5 m/s^2 south

Multiply each side by 20s : V-final = 0.5 m/s^2 south x (20sec) =

10 m/s south

8 0

3 years ago

When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The struc

egoroff_w [7]

Answer:

The structure is shown below.

Explanation:

The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:

FC = V - (L + S/2)

Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.

The possible formulas, from the empiric one, are:

SCl, S₂Cl₂, and S₃Cl₃.

To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:

FC = 7 - (6 + 2/2) = 0

So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)

FC = 6 - (4+ 2/2) = +1

For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so

FC = 6 - (4 + 4/2) = 0

So, it is the correct structure. The lewis structure represents the bonds by lines and the lone pairs of electrons by dots, and it is shown below.

3 0

3 years ago