Compound x has a molar mass of 180.15 g•mol and the following composition: element mass % carbon 40.00% hydrogen 6.71% oxygen 53.29% Write the molecular formula of x.
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- Cyclopropane mixed in the proper ratio with oxygen can be used as an anesthetic. At 755 mm Hg and 25C, it has a density of 1.71 g/L. (a) What is the molar mass of cyclopropane? (b) Cyclopropane is made up of 85.7% C and 14.3% H. What is the molecular formula of cyclopropane?Write the empirical formula for the hydrated KAl(SO4)2, based on moles of anhydrous KSI(SO4)2= 0.046 mol molar mass of H2O= 18g/ mol moles of H2O= 0.0444mol ratio pf moles H20 to moles of anhydrous KAI(SO4)2= 9.65/1 . Show all work including units. Hint: if the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 was 4, then the empirical formula would be: KAl(SO4)2•4H2O.1252. WG compound weighing 1980 h pressure, yielding An unknown organi d 360 mg of water- 12 my s fully combusted at hig ‘9’30 g of carbon dioxide an hat is its empirical formula? A) CHO, B) CHO, ©) CHO, D) CH,0,
- Carbon dioxide emissions associated with a one-night stay in a hotel room are calculated at 28.98 kg of CO2 per room day for an average hotel. The 250 rooms of your hotel are all occupied for two days during a college football game. How much CO2 did the guests and hotel release into the atmosphere? Round your answer to the nearest whole number. kgsQ2: A fuel oil is burned with 65% excess air, and the combustion characteristics of thefuel oil are similar to a paraffin series dodecane C12H26. Determine: (a) the actualair/fuel ratio (A/F) on a mole basis, (b) the actual air/fuel ratio (A/F) on a mass basis,and (c) the actual volumetric (molar) analysis of the products of combustion. State allthe assumptions that you would made before solving this question. Air contains 21%O2 and 79% N2 by mole. Do not solve in tabulated form.The hydrated salt, MY3·xH2O, was heated strongly in a crucible to remove all the watermolecules of crystallization.The data are for this experiment are given in the table below.Mass of empty crucible and cover 29.73 gMass of crucible, cover, and hydrated salt 32.72 gMass of crucible, cover, and anhydrous salt 31.52 gMolar mass of anhydrous salt 241.88 g/mol(i) Calculate the % composition by mass of water in the hydrated salt.(ii) Determine the value of x.(iii) M is in Period 4 of the Periodic Table of the Elements.If Y = NO3–, identify the element M.(iv) What is the name of this hydrated salt? ___
- General formula of a polymer is-(CH2CHCHCN)n-, where n is typically greater than 10,000. Then take a sample of this polymer weighs 755.9 g and contains 3.112 x 1020 molecules of -(CH2CHCHCN)n-, so calculate n?Determination of the amount of water, mass of crucible + dehydrated sample 16.2265g, mass of dehydrated Sample is 1.3031g, mass of water is 0.3587g, moles of water is 0.01991moles,DETERMINATION of the amount of copper, mass of copper 0.6291g, moles of copper 9.900 × 10 to the negative 3 moles, Determination of the amount of chlorine, calculate mass of copper + water.I know the answer is D, but I dont know why. Can you cleave anywhere on the molecule, or only on certain areas? not certain of the rules. Is abundance the m/z (molar mass) of the molecule? I calculated 102 for the molar mass , but that is not one of the options to choose as an answer. Please explain plainly
- Calculate the ratio between the moles of magnesium used and the moles of oxygen used. Express this ratio in simplest whole-number form. Mass of oxygen = 0.49g- 0.33g= 0.16gNumber of moles = given mass / molar massNumber of moles of oxygen = 0.16/ 16 = 0.01 molMolarmass of magnesium = 24.3gMoles of magnesium = 0.33g / 24.3g = 0.0135 molMoles of magnesium 0.0135 1.35 3 4------------------------ = _____________ = ____________ x ________ = ___Moles of oxygen 0.01 1 3 3 The ratio between the moles of magnesium used and the moles of oxygen uses is 4/3. Based on your experimental data, write the empirical formula for magnesium oxide. Is this the molecular formula or the simplest formula? Explain how you know this.Given Active Ingredient: precipitated sulfur (ointment) Raw Materials: 500 g calcium polysulphide and 1.5 kg hydrochloric acid Actual Yield: 343.4g precipitated sulfur Formulation: 250 mg per jar Dosage form: Ointment packaging:100 jars per box Synthesis and Packaging (Need answer)- Balanced Chemical Equation:- % composition by mass of each compound:- Mass to Mass Stoichiometry Calculation:- Limiting Reagent:- Excess Reagent:- Amount (g) in excess: % Yield:- Number of dosage form and packaging that can be produced from stoichiometric solution:Consider a blend of mass 5g formed from 2.2 g of C12H24 and 2.8g of C998H1000 paraffin's:, What are Mn and Mw of the blend?