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The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
Answer:
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Answer:
The molar mass of lysine using the ideal gas equation for this problem is 146.25 g/mole.
Explanation:
The ideal gas equation PV = nRT, was derived from the ABC laws (Avogadros, Boyles and Charles laws). We need to obtain the value for the number of moles n.
The parameters of this equation are:
P = 1.918 atm
V = 750.0mL = 0.75L
n = ?
R = 0.0821
T = 25 degree celcius = 25 + 273 = 298 degree kelvin.
From this formular, n = (PV)/(RT)
n = (1.918 X 0.75)/(0.0821 X 298 )
n = 0.0588
n, no of mole = mass/molar mass
0.0588 = 8.6/MM
MM = 8.6/0.0588
MM = 146.25g/mole.
If the dehydration reaction of an alcohol is successful. The changes would be seen in the IR spectrum for the product compared to the starting material are as,
- The O-H and C-O band is disappear from stating material
- The addition of a C-C double bond band in the product.
In dehydration reaction of alcohol ( O-H and C-O bond ) contain , the water molecule ( 
) is release from the reactant and C-C double bond is form which is known as alkene in the product .
The reactant and product have different structure. To determine the structure of the compound IR spectroscopy is used. In IR spectrum the peak corresponds to 3400-3600 cm is missing in the product of dehydration reaction of an alcohol. It means O-H band is disappear from stating material.
learn about IR SPECTRUM
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