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snow_tiger [21]
3 years ago
9

What is the correct formula for xenon hexafluoride

Chemistry
2 answers:
Andrews [41]3 years ago
5 0
The formula for Xenon Hexafluoride is XeF6
LiRa [457]3 years ago
4 0
Formula for Xenon hexafluoride is XeF6
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Which set of procedures and observations indicates a chemical change?
Flauer [41]
The answer is (4), (3) is merely a change in composition, thus not chemical but physical.

Answer: (4).
4 0
3 years ago
The reaction ch4 (g) + 2 o2 (g) ? co2 (g) + 2 h2o (g) is: the reaction ch4 (g) + 2 o2 (g) co2 (g) + 2 h2o (g) is: an exothermic
barxatty [35]
All of the above
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5 0
3 years ago
a flask was filled with SO2 at a partial pressure of 0.409 atm and O2 at a partial pressure of 0.601 atm. The following gas-phas
jarptica [38.1K]

Answer:

The equilibrium partial pressure of O2 is 0.545 atm

Explanation:

Step 1: Data given

Partial pressure of SO2 = 0.409 atm

Partial pressure of O2 = 0.601 atm

At equilibrium, the partial pressure of SO2 was 0.297 atm.

Step 2: The balanced equation

2SO2 + O2 ⇆ 2SO3

Step 3: The initial pressure

pSO2 = 0.409 atm

pO2 = 0.601 atm

pSO3 = 0 atm

Step 4: Calculate the pressure at the equilibrium

pSO2 = 0.409 - 2X atm

pO2 = 0.601 - X atm

pSO3 = 2X

pSO2 = 0.409 - 2X atm = 0.297

 X = 0.056 atm

pO2 = 0.601 - 0.056 = 0.545 atm

pSO3 = 2*0.056 = 0.112 atm

Step 5: Calculate Kp

Kp = (pSO3)²/((pO2)*(pSO2)²)

Kp = (0.112²) / (0.545 * 0.297²)

Kp = 0.261

The equilibrium partial pressure of O2 is 0.545 atm

3 0
3 years ago
Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole
kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ \frac{[CH_{3}COO^-]}{[CH_{3}COOH]}

a) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[2 mol]}

<em>pH = 4,75</em>

<em></em>

b) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[1mol]}

<em>pH = 5,05</em>

<em></em>

I hope it helps!

7 0
4 years ago
CH, +<br> ,<br> 0, → _CO, +<br> Н,0<br> Balance chemical equation
Flura [38]

Answer:

2 CH2 + 3 O2 = 2 CO2 + 2 H2O

Explanation:

This is what I think that you meant by the question listed. When balancing a chemical equation, you want to make sure that there are equal amounts of each element on each side.

Originally, the equation's elements looked like this: 1 C on left & 1 C on right; 2 H on left & 2 H on right; 2 O on left and 3 O on right. Because these are not balanced, you need to add coefficients.

When adding coefficients, you need to make sure that all of the elements stay balanced, not just one that you are trying to fix. I know that some equations are really difficult to balance, and when that is the case, there are equation balancing websites that can help out.

However, what always helps me is making a chart and continuing to keep up with the changes I am making. It is a trial and error process.

6 0
3 years ago
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