1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Paul [167]
4 years ago
9

The mass of a flask with a 1L volume that is evacuated to a pressure of 0.00 atm is found to be 92.01g. after introduction of an

unknown ideal gas with a pressure of 3.00 atm and temprature of 27 Degrees C the mass of the flask and gas are found to be 97.37g. Identify the unknown gas.
Chemistry
1 answer:
nexus9112 [7]4 years ago
8 0

Answer:

CO₂

Explanation:

We substract the mass of the empty flask from the mass of the flask + gas in order to <u>calculate the mass of the gas</u>:

  • 97.37 g - 92.01 g = 5.36 g

The <em>moles of this gas (n) </em>can be expressed as:

  • n = 5.36 g ÷ M

Where M is the molar mass.

We can <u>calculate n using PV=nRT</u>:

  • P = 3.00 atm
  • V = 1 L
  • T = 27 °C ⇒ 27+273.16 = 300.16 K

3.00 atm * 1L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 300.16 K

  • n = 0.122 mol

Now we <u>can calculate the molar mass of the gas</u>:

n = 5.36 g ÷ M

  • 0.122mol = 5.36 g ÷ M
  • M = 43.9 g/mol ≅ 44 g/mol

Thus, the gas is most likely CO₂.

You might be interested in
A 0.630 gram sample of a metal, M, reacts completely with sulfuric acid according to:A volume of 291 mL of hydrogen is collected
Lina20 [59]

Answer:

55.0 g/mol

Step-by-step explanation:

<em>Step 1</em>. Partial pressure of hydrogen

You are collecting the gas over water, so

p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}

p_{\text{H}_{2}} = p_{\text{atm}} - p_{\text{H}_{2}\text{O}}

p_{\text{atm}} = \text{756.0 Torr}

At 25 °C,p_{\text{H}_{2}\text{O}} = \text{23.8 Torr}

p_{\text{H}_{2}} = \text{756.0 Torr} - \text{23.8 Torr} = \text{732.2 Torr}

===============

<em>Step 2</em>. Moles of H₂

We can use the <em>Ideal Gas Law</em>.

pV = nRT                                       Divide both sides by RT

n = (pV)/(RT)

p = 732.2 Torr                               Convert to atmospheres

p = 732.2/760  

p = 0.9634 atm

V = 291 mL                                     Convert to litres

V = 0.291 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 25 °C                                        Convert to kelvins

T = (25 + 273.15 ) K = 298.15 K    

n = (0.9632 × 0.291)/(0.082 06 × 298.15)      

n = 0.2804/24.47

n = 0.011 46 mol

===============

<em>Step 3</em>. Moles of metal

The partial chemical equation is

M + H₂SO₄ ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.011 46× 1/1

Moles of M = 0.011 46 mol M

===============

<em>Step 4</em>. Atomic mass of M

Atomic mass = mass of M/moles of M

Atomic mass = 0.630/0.011 46

Atomic mass = 55.0 g/mol

7 0
4 years ago
Plssssssssssssssssssssss helppppppppppppppppppppppppppppppppppppp
andriy [413]

Answer:

protons and neutrons have different charges but they have approximately the same mass

3 0
2 years ago
One city is located north of the equator and experiences average rainfall and warm temperatures. Another city is located exactly
antoniya [11.8K]
B I think ..........................
3 0
4 years ago
Which one of the following is an oxidation-reduction reaction?
kupik [55]

Answer:

CH4 + 2 O2 --> CO2 + 2 H2O

Explanation:

CH4 + 2 O2 --> CO2 + 2 H2O is the only reaction where an element (oxygen) undergoes a change in oxidation state. In this reaction oxygen changes disproportionately to O⁻². That is ...

O₂ → CO₂ + 4e⁻ ==> oxidation

<u>O₂ + 4e⁻  →  H₂O ==> reduction </u>

2O₂ + 4e⁻  →  CO₂ + H₂O + 4e⁻  ==> Net oxidation-reduction

=>  4e⁻ gained by one mole O₂ in formation of CO₂ = 4e⁻ lost by the other mole O₂ in forming H₂O.

Then...

Including CH₄ (whose elements do not undergo changes in oxidation states) requires doubling reaction to balance by mass thus giving ...

2CH₄ + 2O₂ + 8e⁻  →  2CO₂ + 2H₂O + 8e⁻

Cancelling 8 reduction electrons on left with 8 oxidation electrons on right gives...

2CH₄ + 2O₂  →  2CO₂ + 2H₂O

7 0
4 years ago
Read 2 more answers
Which requires more heat to warm from 22.0°C and 85.0°C, 45.0 g of water or 200. g of aluminum metal?
stich3 [128]

Answer:

The answer to your question is Aluminum

Explanation:

We need to warm from 22°C to 85°C

a) 45 g of water                      

b) 200 g of aluminum             Cp = 0.905 J/cal°C

a) Water

Q = mCpΔT

Q = (45)(1)(85 - 22) = 45(63) = 2835 cal

b) Aluminum

Q = (200)(0.905)(85 - 22) = 114030 cal

5 0
3 years ago
Other questions:
  • What are the spectator ions in the reaction:
    12·1 answer
  • Determine the name of the compound Li2O
    6·1 answer
  • What is the correct name of this covalent compound BF3?
    15·1 answer
  • Does my profile pic attract females
    15·1 answer
  • You add two substances together in a sealed bag and a chemical reaction happens. The mass before the reaction was 167 g. What wa
    14·1 answer
  • 1.23 grams to pounds
    5·1 answer
  • Oxygen is the ________ component of the atmosphere in terms of volume
    8·1 answer
  • What is the mass of 2.00L of a solution with density of 1.15g/mL?
    13·1 answer
  • Why do you need microscope or hand lens to see the parts that make up-1
    7·2 answers
  • Need help ASAP <br> Dgdbdb
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!