Answer:
6.07 L
Explanation:
It appears that the reading has been made at constant pressure .
At constant pressure , the gas law formula is
V/T = constant V is volume and T is temperature of the gas.
V₁ / T₁ = V₂ / T₂
V₁ = 6.6 L ,
T₁ = 40°C
= 273 + 40
= 313 K
T₂ = 15+ 273
= 288K
V₂ = ?
Putting the values in the formula above
6.6 / 313 = V₂ / 288
V₂ = 6.07 L.
2 C2H6 + 7 O2 ➡️ 4 CO2 + 6 H2O
To identify why a metal measurement was different in the experiments look for the variable that was different in the experiment and analyze how this change affected the results.
<h3>What is an experiment?</h3>
An experiment is a procedure that aims at probing or discovering something. For example, you can test if a plant grows faster/slower by using an experiment.
<h3>What causes different results in similar experiments?</h3>
The most common cause for this situation is that one of the factors or variables is slightly different. For example, if I add 50mL of water to a plant rather than 20mL of water every day this might cause different results.
Based on this, if the metal content was different you should analyze if any of the factors changed in this experiment and find out how this change affected the general results.
Note: This question is incomplete because there is limited information about the experiment; due to this, I answered it based on general knowledge.
Learn more about experiments in: brainly.com/question/13270830
Answer:
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
Explanation:
Step 1: Data given
A sample of sulfur hexafluoride gas occupies a volume of 5.10 L
Temperature = 198 °C = 471 K
The volume will be reduced to 2.50 L
Step 2 Calculate the new temperature via Charles' law
V1/T2 = V2/T2
⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L
⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K
⇒with V2 = the reduced volume of the gas = 2.50 L
⇒with T2 = the new temperature = TO BE DETERMINED
5.10 L / 471 K = 2.50 L / T2
T2 = 2.50 L / (5.10 L / 471 K)
T2 = 230.9 K = -42.1
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
