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nataly862011 [7]
3 years ago
12

7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?

Chemistry
1 answer:
Illusion [34]3 years ago
3 0

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

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koban [17]

Answer:

The answer to your question is 0.67 moles of Glucose

Explanation:

Data

120 g of Glucose

convert to moles

Process

1.- Calculate the molar mass of Glucose

C₆H₁₂O₆ = (12 x 6) + (1 x 12) + (16 x 6)

              = 72 + 12 + 96

              = 180 g

2.- Remember that the molar mass of each molecule equals 1 mol. Use the rule of three to find the answer.

                        180 g ------------------- 1 mol

                         120 g ------------------ x                    

                         x = (120 x 1)/180

                         x = 120/180

                         x = 0.67 moles

3.- Conclusion

120 g of Glucose are equal to 0.67 moles.

8 0
3 years ago
What is the empirical formula for a molecule containing 46.8% Si and 53.2% O?
Over [174]
Assume there are 100 g of the substance.
Masses:
Si: 46.8 g
O: 53.2 g
Moles:
Si: 46.8 / 28 = 1.67
O: 53.2 / 16 = 3.32
Dividing by the smaller number:
Si: 1.6 7/ 1.67 = 1
O: 3.32 / 1.67 = 2
Thus, the formula:
SiO2
4 0
3 years ago
The average American is looking at their phone for 6.25 hours a day. How many hours will the average American spend on their pho
julia-pushkina [17]

Answer:

177253.125hrs

Explanation:

Given parameters:

  Number of hours spend per day on phone  = 6.25hrs

 Unknown:

Average hours pend by an average American if they lived for 77.7yrs

Solution:

Let us find the number of hours an average American spends on their phone per year;

    there are 365days in a year

 Number of hours  = 365 x 6.25  = 2,281.25hrs

Now, if they lived for 77.7yrs;

  They will spend  = 77.7 x 2,281.25 = 177253.125hrs

8 0
2 years ago
What happens when particles vibrate?
dusya [7]
<span>vibration of particles decreases as the temperature decreases It also decreases during phase change but temperature does not</span>
7 0
3 years ago
Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8). 2. What is the pH of a solution containing a hydrogen ion concentrat
Pepsi [2]

Answer:

\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}

Explanation:

1. Solubility of CaF_2

(a) Molar solubility

CaF₂ ⇌ Ca²⁺ + 2F⁻

K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}

(b) Mass solubility

\text{Solubility} = 2.2 \times 10^{-3} \text{ mol/L} \times \dfrac{\text{78.07 g}}{\text{1 L }} = \text{0.17 g/L}\\\\\text{The solubility of CaF$_{2}$ is $\large \boxed{\textbf{0.17 g/L}}$}

2. pH

pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52

3. Oxidizing and reducing agents

Zn + Cl₂ ⟶ ZnCl₂

\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}

The oxidation number of Cl has decreased from 0 to -1.

Cl has been reduced, so Cl is the oxidizing agent.

4. Oxidation numbers

(a) Al₂O₃

\stackrel{\hbox{$\mathbf{+3}$}}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}

1O = -2; 3O = -6; 2Al  = +6; 1Al = +3

(b) XeF₄

\stackrel{\hbox{$\mathbf{+4}$}}{\hbox{Xe}}\stackrel{\hbox{-1}}{\hbox{F}_{4}}

1F = -1; 4F = -4; 1 Xe = +4

(c) K₂Cr₂O₇

\stackrel{\hbox{${+1}$}}{\hbox{K}_{2}}\stackrel{\hbox{$\mathbf{+6}$}}{\hbox{Cr}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{7}}

1K = +1; 2K = +2; 1O = -2; 7O = -14

+2 - 14 = -12

2Cr = + 12; 1 Cr = +6

8 0
3 years ago
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