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Q: What are the specs of the TaylorMade REAX SuperFast 49g?
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What are the number of molecules in 49g of H2SO4?

To find the number of molecules in 49 g of H2SO4, first calculate the molar mass of H2SO4 which is 98 g/mol. Next, use Avogadro's number (6.022 x 10^23) to convert grams to molecules. This gives you approximately 3 x 10^23 molecules in 49g of H2SO4.


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How many lbs and oz equals 49 oz?

1 pound = 16 ounces2 pounds = 32 ounces3 pounds = 48 ounces..10 pounds = 160 ounces..30 pounds = 480 ounces..40 pounds = 640 ounces..49 pounds = 784 ounces


Which solution is more concentrated 1 molarity or 1 normality solution of H2SO4?

A 1 molarity solution of H2SO4 is more concentrated than a 1 normality solution. Molarity is the concentration of a solution based on the number of moles of solute per liter of solution, while normality is the concentration based on the number of equivalents of solute per liter of solution. Since H2SO4 has two acidic protons, its normality is double its molarity.


Method for preparation of 1 normal sulfuric acid?

To prepare 1 normal sulfuric acid (1N H2SO4), you would first need to calculate the molarity of your sulfuric acid solution. Since sulfuric acid is a diprotic acid, each mole of H2SO4 contributes 2 moles of H+ ions. Therefore, to prepare a 1N solution, you would need to dissolve 1 equivalent of H2SO4 in enough water to make 1 liter of solution. The molarity would depend on the concentration of your sulfuric acid solution.


How calculate equivalent weight?

Equivalent Weight is an archaic name for 'Moles'. Moles = mass(g) / Mr ( Relative molecular mass). or Ar (Relative Atomic mass). If we take the reaction of sulphuric acid and sodium hydroxide. First write down the BALANCED reaction eq'n. H2SO4 + 2NaOH = Na2SO4 + 2H2O Notice the molar ratios are 1:2::1:2 So if we have 49g (H2SO4) , how much , by mass of sodium hydroxide, do we need to neutralise the acid. ? This mass of sodium hydroxide is the Equivalent Weight. First calculate the Mr(H2SO4) from atomic masses on the Periodic Table. 2 x H = 2 x 1 = 2 1 x S = 1 x 32 = 32 4 x O = 4 x 16 = 64 2 + 32 + 64 = 98 So moles(H2SO4) = 49g / 98 = 0.5 mole. By molar ratios above 1 mole reacts with 2 moles Hence 0.5 moles reacts with 1 mole. So we need 1 mole(NaOH) Again calculate the Mr(NaOH) 1 x Na = 1 x 23 = 23 1 x O = 1 x 16 = 16 1 x H = 1 x 1 = 1 23 + 16 + 1 = 40 Using the moles equation again moles = mass( g)/Mr Algebraically rearranging mass(g) = moles X Mr Mass(NaOH) = 1 moles X 40 = mass(NaOh) = 40 g ( The 'Equivalent Weight). This is the mass required to neutralise 49 g of sulphuric acid.