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stiv31 [10]
3 years ago
14

Calculate the molarity of a solution prepared by dissolving 13.0 g of Na2CrO4 in enough water to produce a solution with a volum

e of 800. mL .
Express the molarity to three significant digits.
Chemistry
1 answer:
Lelechka [254]3 years ago
8 0

Answer:

molar mass sodium chromate = 162g

12.6g ÷ 162g/mole = moles sodium chromate (This is the solute)

molalrity or M = moles solute ÷ lliters of solution (You'll have to change given mL to L)

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Describe the process of organic vegetable production.​
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4 0
3 years ago
2. 10.00 grams of a sample of hydrated PtCl4 are heated and lose 3.00 g of water. How many moles of water are combined with each
In-s [12.5K]

Answer:

8 mol

Explanation:

Step 1: Calculate the mass of PtCl₄ in the sample

10.00 grams of a sample of hydrated PtCl₄ are heated and lose 3.00 g of water. The mass of PtCl₄ is:

mPtCl₄ = 10.00 g - 3.00 g = 7.00 g

Step 2: Calculate the moles corresponding to 7.00 g of PtCl₄ and 3.00 g of H₂O

The molar mass of PtCl₄ is 336.9 g/mol.

7.00 g × 1 mol/336.9 g = 0.0208 mol

The molar mass of H₂O is 18.02 g/mol.

3.00 g × 1 mol/18.02 g = 0.166 mol

The molar ratio of H₂O to PtCl₄ is:

0.166 mol H₂O/0.0208 mol PtCl₄ ≈ 8 mol H₂O/ 1 mol PtCl₄

8 0
3 years ago
Calculate the weight percent of ascorbic acid in a tablet of Vitamin C from the following data:A 80 mg sample of a crushed Vitam
Lina20 [59]

Answer:

The Answer is 88%

Explanation:

The balanced ionic equation of the reaction

   IO^{3- }+ 8 I^- + 6 H^+ => 3 I^{3- }+ 3 H_2O

  I^{3-} + 2 S_2O_3^{2-} => 3 I^- + S_4O_6^{2-}

 

  C_6H_8O_6 + I^{3- }+ H_2O => C_6H_6O_6 + 3 I^- + 2 H^+

Looking at the above reactions

   The original number of moles of I^{3-} = 3 × number of moles of IO^{3-}

                                                                = 3 × volume × concentration of KIO_3

Note: The formula for number of moles is volume × concentration

                                                                =3 * \frac{40}{1000} *  0.00653 =0.000784 mol

The number of moles of I^{3-} left after its reaction with ascorbic  acid

                         = \frac{1}{2} x moles of S_2O_3^{2-}

                          = \frac{1}{2} x volume x concentration of    S_2O_3^{2-}

                          = \frac{1}{2}  * \frac{15}{1000} * 0.0510 =0.000383 \ mol

Note: The division by 1000 is to convert mill liter to liter

                             Moles of ascorbic acid = moles of I^{3-} reacted

                           = initial \ moles \ of \  I^{3-} \ - remaining \ moles \ of \ I^{3-}

                           =0.000784 - 0.000383

                          =0.000401 \ mols

              Mass =  moles \ * molar \ mass

Hence

              Mass of ascorbic acid = moles of ascorbic acid × molar mass of ascorbic acid

                                                      = 0.000401* 176.12 = 0.07055\ g

                                                      = 70.55\  mg

 Weight% of ascorbic acid = mass of ascorbic acid/mass of sample x 100%

                                             = 70.55/80  × 100%

                                              = 88.1%

                       

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4 years ago
Which element is classified as a noble gas?
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Answer:

Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)

Explanation:

I found my answer on google

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6 0
3 years ago
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