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Fudgin [204]
3 years ago
9

How would you prepare 100.0 ml of.400 m CaCl2 from a stock solution of 2.00 M CaCl2?

Chemistry
2 answers:
exis [7]3 years ago
7 0

Answer is: add 80.00 mL of water to 20.00 mL of stock solution.

V₁(CaCl₂) = ? mL; initial volume of calcium chloride solution.

c₁(CaCl₂) = 2.00 M; initial concentration.

V₂(CaCl₂) = 100.0 mL; final volume.

c₂(CaCl₂) = 0.400 M; final concentration.

Use c₁V₁ = c₂V₂.

2.00 M · V₁(CaCl₂) = 0.400 M · 100.0 mL.

V₁(CaCl₂) = (0.400 M · 100.0 mL) ÷ 2.00 M.

V₁(CaCl₂) = 20 ml.

V(H₂O) = 100 mL - 20 mL.

V(H₂O) = 80 mL.

docker41 [41]3 years ago
4 0
C₀=2 mol/l
c₁=0.400 mol/l
v₁=100.0 ml = 0.1 l

c₁v₁=c₀v₀

v₀=c₁v₁/c₀
v(H₂O)=v₁-v₀

v₀=0.1*0.400/2=0.02 l = 20 ml
v(H₂O)=100 - 20 = 80 ml

It is necessary to mix 20 ml of the feed solution and 80 ml of water.

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Answer:

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Explanation:

7 0
2 years ago
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nikitadnepr [17]

Answer:

a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction

Explanation:

<em>Ferrous Sulphate</em>(FeSO4)<em> is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide </em>(SO_3)<em>, Sulphur Dioxide </em>(SO_2)<em>  as well as Ferric Oxide </em>(Fe_2O_3)<em>.</em>

<em>We can hence, frame a skeletal equation of this reaction and try to balance it.</em>

<em>Hence,</em>

FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3

<em>Now,</em>

<em>a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of </em><em>steps</em><em>:</em>

<em>1. First, lets compare the number of  Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside </em>FeSO_4.

<em>2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:</em>

<em> Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :</em>

<em> </em>2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3

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7 0
2 years ago
Why amino acids are soluble in pH=3 or pH=10 solutions more than it dissolves in pH=7 solutions?
Ray Of Light [21]

Answer:

Solubility is Affected by pH

The pH of an aqueous solution can affect the solubility of the solute. By changing the pH of the solution, you can change the charge state of the solute.

At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

An isoelectric point is the pH at which an amino acid exists as its zwitterion. A zwitterion is the dipolar ionic form of an amino acid. ... If the pH is lower (in acidic conditions) than the isoelectric point then the amino acid acts as a base and accepts a proton at the amino group. This gives it a positive change.

An amino acid is usually more soluble in aqueous solvent at pH extremes than it is at a pH near the isolelectric point of the amino acid. (Note that this does not mean that the amino acid is insoluble at a pH near its pI.)

Which of the following statements correctly explains this phenomenon?

(Select all that apply.)

The neutral charge of an amino acid molecule at its isoelectric point will make the molecule hydrophobic.

At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.

At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

At pH values far from the isoelectric point, individual amino acid molecules have greater kinetic energy, thus more readily stay in solution.

<h2>Please mark me as brainliest</h2>

8 0
3 years ago
Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO;?
faltersainse [42]

Answer:

K₂CO₃    

Explanation:

Given parameters:

Number of moles of K = 0.104mol

Number of moles of C = 0.052mol

Number of moles of O = 0.156mol

Method

From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.

Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.

  • Divide the given moles through by the smallest which is C, 0.052mol.
  • Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
  • This is the empirical formula

Solution

Elements                             K                       C                    O

Number of moles            0.104                0.052            0.156

Dividing by the

smallest                       0.104/0.052     0.052/0.052  0.156/0.052

                                            2                           1                     3

The empirical formula is K₂CO₃      

3 0
2 years ago
Which component of solution is dissolving medium? a.solute b.solvent​
liubo4ka [24]

Answer:

Solvent

Explanation:

7 0
3 years ago
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