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Serjik [45]
2 years ago
9

Quick Help 10 points

Chemistry
2 answers:
Alenkasestr [34]2 years ago
6 0
D - matter remains constant in any chemical equation, so reactants always equal products.
Savatey [412]2 years ago
6 0

Answer:

The products will have one S and four O atoms.

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A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



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3 years ago
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Draw the organic product of the following nucleophilic substitution reaction. Include all hydrogens atoms.
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Is a computer screen energy transfer or transformation?
yarga [219]
A computer screen is energy transfer
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A chemist prepares a solution of sodium thiosulfate by measuring out of sodium thiosulfate into a volumetric flask and filling t
marshall27 [118]

Mass of sodium thiosulfate Na_{2}S_{2}O_{3} is 110. g

Volume of the solution is 350. mL

Calculating the moles of sodium thiosulfate:

110. g Na_{2}S_{2}O_{3} * \frac{1 mol Na_{2}S_{2}O_{3}}{158.1 g Na_{2}S_{2}O_{3}} = 0.696 molNa_{2}S_{2}O_{3}

Converting the volume of solution to L:

350. mL * \frac{1 L}{1000 mL} = 0.350 L

Finding out the concentration of solution in molarity:

\frac{0.696 mol}{0.350 L} =  1.99 mol/L

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The percent composition of calcium is ?
DochEvi [55]

Answer: 40.1%

Explanation: The mass of calcium in this compound is equal to 40.1 grams because there's one atom of calcium present and calcium has an atomic mass of 40.1 . The molar mass of the compound is 100.1 grams. Using the handy equation above, we get: Mass percent = 40.1 g Ca⁄100.1 g CaCO3 × 100% = 40.1% Ca.

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