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Akimi4 [234]
3 years ago
6

Changing _________________ can significantly alter the area that could be affected by a chemical release.

Chemistry
1 answer:
Harman [31]3 years ago
6 0
The answer is D, location, because you'd have moved spots that could be altered.
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The equilibrium constant, Kp, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g) 2CH2Cl2(g) Calculate the equili
dimaraw [331]

Answer:

pCH4 =  0.9184 atm

pCCl4 = 0.9184 atm

pCH2Cl2 = 0.2832 atm

Explanation:

Step 1: Data given

The equilibrium constant, Kp= 9.52 * 10^-2

Temperature = 350 K

Each have an initial pressure of 1.06 atm

Step 2: The balanced equation

CH4(g) + CCl4(g) ⇆ 2CH2Cl2(g)

Step 3: The pressure at the equilibrium

pCH4 = 1.06 - X atm

pCCl4 = 1.06 - X atm

pCH2Cl2 = 2X

Step 4: Calculate Kp

Kp = (2X)² / (1.06 - X)*(1.06 - X)

9.52 * 10^-2 = 4X² / (1.06 - X)*(1.06 - X)

X = 0.1416

Step 5: Calculate the partial pressure

pCH4 = 1.06 - 0.1416 =  0.9184 atm

pCCl4 = 1.06 - 0.1416 =  0.9184 atm

pCH2Cl2 = 2 * 0.1416 = 0.2832 atm

Kp = (0.2832²) / (0.9184*0.9184)

Kp = 9.52 * 10^-2

pCH4 =  0.9184 atm

pCCl4 = 0.9184 atm

pCH2Cl2 = 0.2832 atm

3 0
3 years ago
six different aqueous solutions are represented in the beakers below, and there total volumes are noted. (a) which solution has
wolverine [178]

The molarity of the solutions are as follows:

  • solution B has the highest molarity
  • solutions A, D and F have the same molarity
  • solutions A and C are mixed together have a lower molarity than B
  • solution F and D will have the same molarity
  • Volume of water required to be evaporated is 8.3 mL

<h3>What is molarity of a solution?</h3>

The molarity of a solution is the amount in moles of a substance present in a given volume of solution.

From the image of the solution given:

  • solution B has the highest molarity
  • solutions A, D and F have the same molarity
  • when solutions A and C are mixed, the resulting solution have a lower molarity than B
  • solution F and D will have the same molarity after 75 mL and 50 mL of water are added to each respectively
  • the molarity of B is 12/50 = 4/16.7. Volume of water required to be evaporated = 25 - 16.7 = 8.3 mL

Therefore, the molarity of the solutions depends on the moles of substance present per given volume of solution.

Learn more about molarity at: brainly.com/question/24305514

#SPJ1

6 0
2 years ago
I need help with this question
denis-greek [22]
It’s B. I had that to
6 0
4 years ago
Use Boyle's Law to solve this problem: A gas has a volume of 10.0 L at a pressure of 4.0 atm. If the gas expands to 20.0 L, what
NARA [144]

Answer:

P₂ = 2 atm

Explanation:

Given data:

Initial volume = 10.0 L

Initial pressure = 4.0 atm

Final volume = 20.0 L

Final pressure = ?

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

4.0 atm × 10.0 L = P₂ × 20.0 L

P₂ = 40.0  atm. L/ 20.0 L

P₂ = 2 atm

7 0
3 years ago
La columna de la izquierda corresponde a los tipos de sales y la columna derecha a los tipos de fórmula que presentan. Relaciona
Galina-37 [17]

Answer:

1. Hidracidas a. MX

2 Acidas c. MHXO

3. Oxacidas  b. MXO

4. Basicas d. M(OH)X

Explanation:

¡Hola!

En este caso, de acuerdo con el concepto de sal, la cual está generalmente dada por la presencia de al menos un metal y un no metal, es posible encontrar cuatro tipos de estas; hidrácidas, oxácidas, básicas y ácidas, en las que las primeras dos son neutras pero la segunda tiene presencia de oxígeno, la tercera tiene iones hidróxido adicionales y la cuarta iones hidrógeno de más.

Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:

1. Hidracidas a. MX

2 Acidas c. MHXO

3. Oxacidas  b. MXO

4. Basicas d. M(OH)XO

En las que M se refiere a un metal, X a un no metal, H a hidrógeno y O a oxígeno.

¡Saludos!

3 0
3 years ago
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