Amixture of helium and krypton gases at a total pressure of 904 mm hg contains helium at a partial pressure of 718 mm hg. if the gas mixture contains 0.777 grams of helium, how many grams of krypton are present?

If helium has partial pressure of 718 mmHg, the krypton must have (904 mmHg - 718 mmHg = ) 186 mmHg. Since temperature and volme are the same for each gas, the mole ratio of helium to krypton must be 718 / 186. The dimensional analysis then yields:

0.777 g He * 1 mol He / 4.00 g He * 186 mol Kr / 718 mol He * 83.80 g Kr / mol Kr = 4.22 g Kr.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

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