Answer:
Exosphere
Explanation:
it is found at the end reaching outer space
Answer:
The group 18 elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). These elements are non-reactive and are called noble gases as they have their outermost orbit complete. Due to stable electronic configuration they hardly react with other elements.The name comes from the fact that these elements are virtually unreactive towards other elements or compounds.The group 18 elements are called noble or inert gases. As the name suggests these are inert because chemically they are very less reactive or not at all reactive.The full valence electron shells of these atoms make the noble gases extremely stable.
It is in period 3
It is in group 17
It is a chlorine atom because it has 17 electrons which means the atomic number is 17
Answer : The products are Silver sulfide,
and Sodium iodide,
.
Explanation :
The given balanced chemical reaction is,

From the given balanced reaction, we conclude that the 2 moles of silver iodide react with the 1 mole of sodium sulfide to give product as 1 mole of silver sulfide and 2 moles of sodium iodide.
In a chemical reaction, reactants are represent on the left side of the right-arrow and products are represent on the right side of the right-arrow.
Therefore, in a chemical reaction the products are Silver sulfide and Sodium iodide.
CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).