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Andrew [12]
3 years ago
7

The simplest form of all matter are small partials called

Chemistry
1 answer:
iris [78.8K]3 years ago
6 0

All matter is made up of small particles called atoms.

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Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices
lilavasa [31]

Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

  • P= 0.980 arm
  • V= 1.20 L
  • T= 287 K
  • R= 0.082 \frac{atmL}{molK}
  • n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082\frac{atmL}{molK}× 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082\frac{atmL}{molK}× 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

molar mass=\frac{0.458 grams}{0.04997 moles}

Solving:

<u><em>molar mass= 9.17 </em></u>\frac{g}{mol}

Finally, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

Learn more about

molar mass:

brainly.com/question/5216907

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brainly.com/question/7132033

brainly.com/question/17249726

ideal gas law:

brainly.com/question/4147359

#SPJ1

6 0
2 years ago
Using the standard enthalpies of formation found in the textbook, determine the enthalpy change for the combustion of ethanol c2
ArbitrLikvidat [17]
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)

Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol

Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol

<em>Answer: -1367.5 kJ/mol</em>
6 0
3 years ago
Read 2 more answers
Which one is NOT part of the cell theory?
worty [1.4K]

<u>Answer:</u>

All living things are not made of cells.

5 0
2 years ago
This is very difficult
Oliga [24]

Answer: C

Explanation:

6 0
3 years ago
A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres
tatiyna

Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 m^{3}

Mass of the gas = 1.15 gm = 0.00115 kg

Temperature = 25 ° c = 298 K

Gas constant for Argon R = 208.13 \frac{J}{kg k}

From ideal gas equation P V = m RT

⇒ P = \frac{m R T}{V}

Put all the values in above formula we get

⇒ P = \frac{0.00115}{0.001} × 208.13 × 298

⇒ P = 71.326 K pa

Therefore, the partial pressure of argon in the flask = 71.326 K pa

4 0
3 years ago
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