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charle [14.2K]
3 years ago
11

If two gases with pressures of 2 atm and 3 atm are mixed at a constant temperature, what will the total pressure be?

Chemistry
2 answers:
slamgirl [31]3 years ago
8 0
If two gases with pressures of 2 atm and 3 atm are mixed at constant temperature, the total pressure will be the sum of the two pressures. Therefore the answer is D. 2 atm + 3 atm or 5 atm will be the total pressure of the gas mixture. 
kotegsom [21]3 years ago
6 0

Answer:

D. 2 atm + 3 atm

Explanation:

Given:

Pressure Gas 1 = 2 atm

Pressure Gas 2 = 3 atm

Explanation:

As per Dalton's Law in a mixture of gases, the total pressure is the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + P3 + ...

where P1, P2,P3 are the partial pressures

Therefore for the given gas mixture, the total pressure would be:

Ptotal = 2 atm + 3 atm = 5 atm

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Sonja [21]
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The Haber-Bosch process operates at reaction pressures of 200.0
kiruha [24]
<h3>Haber - Bosch process, method of directly synthesizing ammonia from hydrogen... The reaction is carried out at pressure ranging from 200 to 400 atmosphere's</h3>

sana maka tulong ❣️

8 0
3 years ago
2C3H7OH + 9O2 --&gt; 6CO2 + 8H2O
Ne4ueva [31]
<h3>Answer:</h3>

733 g CO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given]   5.55 mol C₃H₇OH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                    \displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})
  2. Multiply/Divide:                                                                                               \displaystyle 732.767 \ g \ CO_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

732.767 g CO₂ ≈ 733 g CO₂

8 0
3 years ago
Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °
Mila [183]
This problem is to use the Claussius-Clapeyron Equation, which is:

ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]

Where p2 and p1 and vapor pressure at estates 2 and 1

ΔH is the enthalpy of vaporization

R is the universal constant of gases = 8.314 J / mol*K

T2 and T1 are the temperatures at the estates 2 and 1.

The  normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa

Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol 

=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]

=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x =  0.000157 + 1/330.95 = 0.003179

=> x = 314.6 K => 314.6 - 273.15 = 41.5°C

Answer: 41.5 °C 
3 0
3 years ago
Lana balanced an equation so that the result was 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr. Which most likely represents the starting e
seropon [69]

Answer:

C2H3Br + O2 → CO2 + H2O + HBr

Explanation:

The term balancing of chemical reaction equation has a unique meaning in chemistry. What it actually means is to ensure that the number of atoms of each element on the left hand side of reaction equation becomes equal to the number of atoms of the same element on the right hand side of the reaction equation.

When we look at the equation; C2H3Br + O2 → CO2 + H2O + HBr, the number of atoms of each element on the left and right hand sides of the given equation are not the same hence the equation is unbalanced.

If we look at the equation; 2C2H3Br + 5O2 → 4CO2 + 2H2O + 2HBr, the number of atoms of each element on both sides of the reaction equation are now equal, thus the later equation is the balanced version of the former.

6 0
3 years ago
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