Prepare ~ 8.75 104 M CrO4 solution from the ~1.25 103 M K2CrO4 stock solution.

Prepare ~ 8.75 × 10–4 M CrO4 solution from the ~1.25× 10–3 M K2CrO4 stock solution. Record the exact molarity of the stock solution. Calculate the required amount of stock solution to make 10.00 mL of dilute solution. Adjust this value as necessary (ratio) based on the volumetric flask available.

2 months ago

Solution 1

Guest Guest #1625861
2 months ago

Answer:

7 mL

Explanation:

To prepare a solution by dilution of another one, we can use the equation:

C1V1 = C2V2

where C is the concentration, V is the volume, 1 is the initial solution, and 2 the final solution. V represents the total volume. K₂CrO₄ dissolver by:

K₂CrO₄ → 2K⁺ + CrO₄⁻²

So, 1 mol of K₂CrO₄ results in 1 mol of CrO₄⁻², the in a solution of 8.75x10⁻⁴M of CrO₄⁻², it will be 8.75x10⁻⁴M of K₂CrO₄, so the volume of the stock solution will be:

1.25x10⁻³xV1 = 8.75x10⁻⁴x10

V1 = (8.75x10⁻³)/(1.25x10⁻³)

V1 = 7 mL

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Question
Consider the equilibrium reaction: 3CIO-(aq) ↔ CIO3-(aq) + 2CI-(aq) The equilibrium constant Kc = 3.2 x 103. The following concentrations are present: [Cl-] = 0.50 mol/L; [ClO3-] = 0.32 mol/L; [ClO-] = 0.24 mol/L. Is the mixture at equilibrium and, if not, in which direction will reaction proceed?
Solution 1

Answer:

Forward direction

Explanation:

The reaction quotient of an equilibrium reaction measures relative amounts of the products and the reactants present during the course of the reaction at  particular point in the time.

Q < Kc , reaction will proceed in forward direction.

Q > Kc , reaction will proceed in backward direction.

Q = Kc , reaction at equilibrium.

It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.

Thus, for the reaction:

3CIO^{-}_{(aq)}\rightleftharpoons CIO_3^{-}_{(aq)}+2Cl^{-}

The expression is:

Q=\frac {[CIO_3^{-}][Cl^{-}]^2}{[CIO^{-}]^3}

Given,

[Cl⁻] = 0.50 mol/L; [ClO₃⁻] = 0.32 mol/L; [ClO⁻] = 0.24 mol/L

So,

Q=\frac{0.32\times (0.50)^2}{(0.24)^3}

Q = 5.7870

Since, Q < Kc (3.2\times 10^3)

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Question
Ain___accepts electrons and becomes___
Solution 1

Answer:

An atom accepts electrons and becomes an anion.

Explanation:

Atomic number = number of electrons = number of protons.

Atomic number of sodium is 11

So the atom contains 11 protons and 11 electrons

To find the number of neutrons we make use of the formula  

Mass number - atomic number = number of neutrons

From the periodic table, we know mass number of sodium is 23

So number of neutron = 23 - 11 = 12.

When a sodium atom loses an electron it will have 11 positive protons and 10 negative electrons. Since 1 positive charge is more, Na becomes Na^+.

Positively charged ion is called as cation

Chlorine's atomic number is 17 so it has 17 protons and 17 electrons.

When it gains an electrons, it will have 17 positive protons and 18 negative electrons. Since 1 negative charge is more, Cl becomes Cl^-.

Negatively charged ion is called as anion.

Question
25 mL of a concentrated solution of sodium chloride is added to a 500 mL volumetric flask and sufficient water added to make up to the mark. The concentration of this diluted solution is 0.125 M. What was the concentration of the original solution?
Solution 1

Answer:

2.5 M

Explanation:

To address this question, we can use the dilution formula:  

M1V1= M2V2

Where M1 is the concentration of the concentrated solution  

V1 is the volume of the concentrated solution (the amount of mL that we                            take from the initial solution)  

        M2 is the concentration of the diluted solution  

V2 is the volume of the diluted solution or the final volume.  

For this case, we know V1, M2 and V2 and we want to know M1.  

Following the formula, M1=  \frac{V2*M2}{V1}

Therefore, M1 =  \frac{(500)*(0.125)}{25}

For this, the  concentration of the original solution is 2.5 M.

I hope this clarifies your inquiry.

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Can somebody please answer these 5 questions for me! Thank you :) In the following problems, practice calculations involving energy, wavelength and frequency: 1) What is the frequency of green light that has a wavelength of 5.14 x 10^-7m? 2) What is the wavelength of infrared radiation with a frequency of 3.85 x 10^12 Hz? 3) What is the energy of a photon with a frequency of 7.85 x 10^14 Hz? 4) What is the energy of a photon having a wavelength of 6.08 x 10^-7 m? 5) What is the wavelength of ultraviolet radian having 2.94 x 10^-8 J of energy?
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Answer:

1. The frequency of green light that has a wavelength of the 5.14 \times 10^-^7 m \ is \ 5.83 \times 10^1^4 Hz.

2. Wavelength of infrared radiation with frequency 3.85 \times 10^1^2  Hz \ is \ 7.79 \times 10^-^5  m.

3. Energy of a photon with frequency 7.85 \times 10^1^5  is \  5.181 \times 10^-^1^8  J.

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5. Wavelength of ultraviolet radiation having 2.94 \times 10^-^8J \ is  \ 6.73 \times 10^-^1^8m.

Explanation:

The equation connecting wavelength, frequency and speed of electromagnetic radiation is

c=ϑλ

1. λ = 5.14 \times 10^-^7m

c=3 \times 10^8 m/s

ϑ = c/λ = \frac{(3 \times ^8)}{(5.14 \times 10^-^7)} = 5.83 \times 10^1^4  Hz

2. ϑ = 3.85 \times 10^1^2  Hz

λ= c/ϑ = \frac{ (3 \times 10^8)}{(3.85 \times 10^1^2 )} = 7.79 \times 10^-^5  m

3. E = h ϑ

= 6.6 \times 10^-^3^4 \times 7.85 \times 10^1^5=5.181 \times 10^-^1^8J

4. E= hc/λ

= \frac{ (6.6 \times 10^-^3^4 \times 3 \times 10^8)}{ (6.08 \times 10^-^7)}

=3.25 \times 10^-^1^9J

5. E=2.94 \times 10^-^8J

λ= hc/E

= \frac{ (6.6 \times 10^-^3^4) \times 3 \times 10^8}{(2.94 \times 10^-^8)}

=6.73 \times 10^-^1^8m

Question
Identify the missing coefficient in the balanced equation and classify the type of reaction. 2H3PO4 +3Ba(OH)2 — Baz(PO4)2 + H2O 6; Neutralization 3; Combustion 6; Combustion 3; Neutralization
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Answer:

Option 6 ) Neutralization

Explanation:

For this case, the missing coefficient would be a "6" before the H₂O, within final products (right side of the equiation), hence, the final reaction should be:

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You should have in mind that the amount of atoms at each side of the chemical equation should be the same, so as to comply with the principle of mass conservation. If you add "6" on the left side of the H₂O, the equation will be balanced (for each side, lef and right, you will have: 12H, 2P, 14O and 3Ba)

Lastly, this is a chemical neutralization reaction, where an acid (H₃PO₄) is reacting with a base (Ba(OH)₂) in order to finally obtain a neutral salt (Ba₃(PO₄)₂) and water (H₂O)

Solution 2

Answer:

6; Neutralization

Explanation:

2H_3 PO_4+3Ba(OH)_2>Ba_3 (PO_4)_2+H_2 O

(Unbalanced)

Balancing is making the number of atoms of each element same on both the sides  (reactant and product side).

To find the number of atoms of each element we multiply coefficient and the subscript  

For example 5Ca_1Cl_2 contains  

5 × 1 = 5 ,Ca atoms and

5 × 2 = 10, Cl atoms  

If there is a bracket in the chemical formula  

For example 3Ca_3 (P_1 O_4 )_2 we multiply coefficient × subscript × number outside the bracket to find the number of atoms  

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

So

3 × 3 = 9 , Ca atoms  

3 × 1 × 2 = 6, P atoms  

3 × 4 × 2 = 24, O atoms are present

Reactant side           Number of atoms or groups                   Product side

                                       of each element    

3                                                 Ba                                                           3

2                                                 PO_4                                      2

12                                                H                                                             2

6                                                 O (Exclude O of PO_4)         1

To balance H and O, multiply H_2O by 6 so we have

2H_3 PO_4+3Ba(OH)_2>Ba_3 (PO_4)_2+6 H_2 O

Reactant side           Number of atoms or groups                    Product side

                                       of each element  

3                                                 Ba                                                           3

2                                                 PO_4                                      2

12                                                H                                                            12

6                                                 O (Exclude O of PO_4)        6

Balanced!!!!

Acid and base reacts to form salt and water is called a neutralization reaction.

Question
Identify the missing coefficient in the balanced equation and classify the type of reaction. CasN2 + 6H20 _Ca(OH)2 + 2NH3 4: Single Displacement 3. Single Displacement 4: Double Displacement 3; Double Displacement
Solution 1

Answer:

3; Double Displacement

Explanation:

Balancing is making the number of atoms of each element same on both the sides  (reactant and product side).

To find the number of atoms of each element we multiply coefficient and the subscript  

For example 5Ca_1Cl_2 contains  

5 × 1 = 5 ,Ca atoms and

5 × 2 = 10, Cl atoms  

If there is a bracket in the chemical formula  

For example 3Ca_3 (P_1 O_4 )_2 we multiply coefficient × subscript × number outside the bracket to find the number of atoms  

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

So

3 × 3 = 9 , Ca atoms  

3 × 1 × 2 = 6, P atoms  

3 × 4 × 2 = 24, O atoms are present

Let us balance the equation  given

Ca_3 N_2+6H_2 0>Ca(OH)_2+2NH_3

(UNBALANCED)

Reactant side           Number of atoms or groups                   Product side

                                       of each element    

3                                                 Ca                                                           1

2                                                  N                                                            2

12                                                H                                                            8

6                                                  O                                                            2

Multiplying Ca(OH)_2 on the right side by 3, we will have

Ca_3 N_2+6H_2 0>3Ca(OH)_2+2NH_3

Reactant side           Number of atoms or groups                   Product side

                                       of each element    

3                                                 Ca                                                           3

2                                                  N                                                            2

12                                                H                                                            12

6                                                  O                                                            6

Balanced!!!!

Solution 2

3: double displacement

Question
HELP WILL GIVE BRAINLIESTWhich of these elements would have the largest ionic radius? Bi3+ C4- Li+ Na+ Cl-
Solution 1

Ions can be made by single element or covalently bonded group of elements. The covalently bonded group of elements is called polyatomic ions or polyatomic atoms. Therefore, C⁴⁻ is the largest among all.

What is Ions?

Any species that contain charge whether it is positive charge or negative charge is called ions. The example of polyatomic ions are sulfate, phosphate, nitrate etc.

Cation is the species that loose electron and attain positive charge while anion is a species which gain electron and attains negative charge so when anion and cation combine in fixed ration the the overall charge of the molecule is zero that is molecule is neutral. Size of cation is always smaller than anion. So, C⁴⁻ is the largest among all.

Therefore, C⁴⁻ is the largest among all.

To learn more about ions, here:

brainly.com/question/13692734

#SPJ3

Solution 2

Answer:

Bi3+ has 103

C4- has 170

Li+ 76

Na+102

Cl-185