The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L
From the question given above, the following data were obtained:
Initial pressure (P₁) = 8.5 atm
Initial volume (V₁) = 24 L
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Final pressure (P₂) = 13.5 atm
Final temperature (T₂) = 15 °C = 15 + 273 = 288 K
<h3>Final volume (V₂) =? </h3>
- The final volume of the gas can be obtained by using the combined gas equation as illustrated below:
Cross multiply
298 × 13.5 × V₂ = 204 × 288
4023 × V₂ = 58752
Divide both side by 4023
<h3>V₂ = 15 L </h3>
Therefore, the final volume of the gas is 15 L
Learn more: brainly.com/question/25547148
Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:
With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:
I hope it helps!
Answer:
beryllium iodide has a molar mass of 262.821 g mol−1 , which means that 1 mole of beryllium iodide has a mass of 262.821 g . To find the mass of 0.02 moles of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.
Explanation:
Answer:
15.28 L
Explanation:
Use combined gas law and rearrange formula
Change C to K
- Hope that helped! Please let me know if you need further explanation, as I can show you step by step.
Answer:
0.665 moles of CO₂
Explanation:
The balance chemical equation for the combustion of Ethane is as follow:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
Step 1: <u>Calculate moles of C₂H₆ as;</u>
Moles = Mass / M.Mass
Putting values,
Moles = 10.0 g / 30.07 g/mol
Moles = 0.3325 moles
Step 2: <u>Calculate Moles of CO₂ as;</u>
According to balance chemical equation,
2 moles of C₂H₆ produced = 4 moles of CO₂
So,
0.3325 moles of C₂H₆ will produce = X moles of CO₂
Solving for X,
X = 0.3325 mol × 4 mol ÷ 2 mol
X = 0.665 moles of CO₂
