please help, with CLEAR steps so I can follow along please! 1a. 5.0 g of O2...
please answer both. I just need these two last questions and
there due at midnight.
Oxygen gas can be prepared by heating potassium chlorate according to the following equation 2KCIO3(s) *2KCI(s) + 302(g) The product gas, O2, is collected over water at a temperature of 25 °C and a pressure of 747 mm Hg. If the wet O2 gas formed occupies a volume of 6.91 L, the number of grams of O, formed is g. The vapor pressure of water...
Zinc reacts with hydrochloric acid to form hydrogen gas. A 0.258 g sample of Zinc reacts with 25.0 mL of 0.250 M HCI solution. What volume does the hydrogen has formed occupy at 24.0 °C and 0.990 atm? Zn(s) + 2HCl(aq) --ZnCl2(aq) + H2(g) 33.0 mL 97.3 mL 6.21 mL 77.0 mL
zinc metal reacts with excess hydrochloric acid to produce hydrogen gas according to the following equation: Zn(s) + 2HCl(aq)ZnCl2(aq) + H2(g) The product gas, H2, is collected over water at a temperature of 20 °C and a pressure of 745 mm Hg. If the wet H2 gas formed occupies a volume of 9.32 L, the number of moles of Zn reacted was __mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.
Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) Part A: How many liters of H2 would be formed at 538 mm Hg and 16∘C if 22.0 g of zinc was allowed to react? Part B: How many grams of zinc would you start with if you wanted to prepare 5.65 L of H2 at 350 mm Hg and 34.5 ∘C?
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Zinc reacts with hydrochloric acid according to the reaction equation Zn(s) +2HCl(aq) ? ZnCl2(aq)-H2(g How many milliliters of 4.50 M HCl(aq) are required to react with 5.25 g of an ore containing 50.0% Zn(s) by mass? Number mL
Calculate and use the molar volume of a gas under ndard temperature and pressure conditions Question The Periodic Table of Elements He Rn Fr Ho Er Tm Md No Zinc will react with hydrochloric acid to produce hydrogen gas. Zn(s)+2 HCl(aq) ZnCl2(aq)H2 (g) What is the volume of H2 gas produced if 2.00 g of zinc is used with an excess of hydrochloric acid in the reaction above at STP? Round to 3 significant figures. Provide your answer below: Oal-...
Q2 part1. Hydrogen gas can be readily prepared by reacting zinc metal with hydrochloric acid: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) A sample of zinc metal was decomposed in hydrochloric acid and the hydrogen gas was collected over water. The volume of gas collected is 0.798 L at 25oC and a total pressure of 735 torr. How many grams of zinc were decomposed assuming there is an excess of hydrochloric acid? [Hint: the vapor pressure of water at 25oC...
When 0.40 g of impure zinc reacted with excess hydrochloric acid, 127mL of hydrogen gas were collected over water at 10 degrees C at a total pressure of 737.77 Torr. The vapor pressure of water at 10 degrees C is 9.21 Torr. The reaction in question is: Zn (s) + 2HCl (aq) --> ZnCl2 (aq) + H2 (aq) A.) what amount (in grams) of H2 gas was collected? B.) what is the percentage of purity of the zinc, assuming that...
Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) How many liters of H2 would be formed at 680 mm Hg and 17 ∘C if 27.0 g of zinc was allowed to react? How many grams of zinc would you start with if you wanted to prepare 6.00 L of H2 at 400 mm Hg and 35.0 ∘C?
please do all. thanks. 1. a. A weather balloon is filled to the volume of 105 L on a day when the temperature is 25°C. If no gases escaped, what would be the volume of the weather balloon after it rises to an altitude where the temperature is -9°C? Volume = L? b A 203-mL sample of a gas exerts a pressure of 2.55 atm at 19.5°C. What volume would it occupy at 1.85 atm and 190.°C? Volume = ?mL...