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suter [353]
3 years ago
6

Please help :,) I’m being timed

Chemistry
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

B

Explanation:

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A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f
Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

6 0
2 years ago
Elements that have complete valence electron shells are mostly found in the
nika2105 [10]
Noble gases have complete valence electron shells
6 0
2 years ago
Read 2 more answers
calcium reacts with fluorine to produce calcium fluoride. how does oxidation and reduction take place in this reaction?
Assoli18 [71]

The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.

<u>Explanation:</u>

An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.

So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as c a^{+} ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to F^{-} ions. This process of acceptance of electrons is termed as reduction.

3 0
3 years ago
Using the following equation: 3H2SO4+ 2Fe -&gt; Fe2(SO4)3+ 3H2
Mariana [72]

Answer:

H₂SO₄

Explanation:

Given data:

Number of moles of H₂SO₄ = 15 mol

Number of moles of Fe = 13 mol

Which reactant is limiting reactant = ?

Solution:

Chemical equation:

3H₂SO₄  + 2Fe      →          Fe₂(SO₄)₃  + 3H₂

now we will compare the moles reactant with product.

               H₂SO₄         :          Fe₂(SO₄)₃  

                  3               :              1

                 15               :              1/3×15 = 5

                H₂SO₄         :            H₂

                  3               :              3

                 15               :              15

                Fe               :          Fe₂(SO₄)₃  

                  2               :              1

                 13               :              1/2×13 = 6.5

                Fe               :                H₂

                  2               :                 3

                 13               :              3/2×13 = 19.5

Number of moles of product formed by  H₂SO₄ are less thus it will act as limiting reactant.

8 0
3 years ago
A pharmacist wishes to strengthen a mixture from 10%alcohol to 30% alcohol. How much pure alcohol should be added to 7 liters of
Tomtit [17]

Answer:

2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.

Explanation:

Suppose x is the number of litres added to the 10% mixture than the quantity of new mixture is given as below

  • n_{old}=7 litres
  • n_{new}=7+x litres

Also the quantity of alcohol is given as

  • q_{old}=10 \% \, of \, 7 \, litres =0.7
  • q_{added}=x
  • q_{new}= 30 \% \,of \,new\, quantity = 0.3(7+x)

Now the equation is as

                                  q_{old}+q_{added}=q_{new}\\0.7+x=0.3(7+x)\\0.7+x=2.1+0.3x\\x-0.3x=2.1-0.7\\0.7x=1.4\\x=2 \, litres

So 2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.

8 0
3 years ago
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