Since the container of the gas is rigid, the volume of the gas will remain constant. Therefore, when the number of particles were decreased in half then the pressure will also be half of the original given they both are subjected to the same temperature.
PV = nRT
V, T and R are constants so they can be lumped together to a constant k.
P/n = k
P1/n1 = P2/n2
since n2 = n1/2
P1/n1 = P2/<span>n1/2</span>
P2 = P1/2
Answer:An iodide ion is the ion I−. Compounds with iodine in formal oxidation state −1 are called iodides. This page is for the iodide ion and its salts, not organo iodine compounds. In everyday life, iodide is most commonly encountered as a component of iodized salt, which many governments mandate.
brainliest pls
The answer is B
Explanation:
Can you mark brainliest I correct
To reduce a haloalkane, it has to be eliminated first. This is because 2 bromopropanes are a saturated compound that can not be reduced by a saturated compound. Dehydrohalogenation of the haloalkanes into the form of propene and hydrogen bromide can lead to the elimination. The propene is then reduced to propane.
If you overheat copper sulfate higher of mass will be lost that is copper sulfate will loss sulfur and oxygen which led to a higher loss of mass than if you would have heated enough. This higher mass lost will be shown in calculation as percentage of water lost