Answer:

Explanation:
Hello,
In this case, since you have both the final and initial volume for the titration procedure, clearly, the used volume is computed by the subtraction between them, since it accounts for the employed volume of the base, it turns out:

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:<span> </span><span>4 HCl + MnO2 → MnCl2 + 2 H2O + Cl2
(33.7 g MnO2) / (86.93691 g MnO2/mol) = 0.38764 mol MnO2
(45.3 g HCl) / (36.4611 g HCl/mol) = 1.2424 mol HCl
(a)
1.2424 moles of HCl would react completely with 1.2424 x (1/4) = 0.3106 mole of MnO2, but there is more MnO2 present than that, so MnO2 is in excess and HCl is the limiting reactant.
(b)
(1.2424 mol HCl) x (1 mol Cl2 / 4 mol HCl) x (70.9064 g Cl2/mol) = 22.0 g Cl2
(c)
(77.7% of 22.0 g Cl2) = 17.1 g Cl2</span>
Answer and explanation:
The attached figure shows five different structures for the chemical formula C4H5O5, but only one of these structures represent the malic acid.
Malic acid is a dicarboxylic acid, this means malic acid has two
-COOH groups. Also, malic acid is a secundary alcohol, which means it has a R2-C-OH group.
-Structure A has two carboxylic groups, but it doesn´t have a secundary alcohol.
-Structure B doesn´t have any caborxylic group.
-Structure C has two carboxylic groups and it is a secundary alcohol. Structure C is the Malic acid
<u>Answer:</u> The rate of the reaction when concentrations are changed is 
<u>Explanation:</u>
For the given chemical reaction:

The given rate law of the reaction follows:
![\text{Rate}_1=k\frac{[A][C]^2}{[B]^{1/2}}](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D_1%3Dk%5Cfrac%7B%5BA%5D%5BC%5D%5E2%7D%7B%5BB%5D%5E%7B1%2F2%7D%7D)
We are given:

When the concentrations are changed:
New concentration of A = 2A (Concentration is doubled)
New concentration of C = 2C (Concentration is doubled)
New concentration of B = 3B (Concentration is tripled)
The new rate law expression becomes:
![\text{Rate}_2=k\frac{[2A][2C]^2}{[3B]^{1/2}}\\\\\text{Rate}_2=\frac{2\times 2^2}{3^{1/2}}\times (k\frac{[A][C]^2}{[B]^{1/2}})\\\\\text{Rate}_2=4.62\times (\text{Rate}_1)\\\\\text{Rate}_2=4.62\times 1.12\times 10^{-2}\\\\\text{Rate}_2=5.17\times 10^{-2}M/s](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D_2%3Dk%5Cfrac%7B%5B2A%5D%5B2C%5D%5E2%7D%7B%5B3B%5D%5E%7B1%2F2%7D%7D%5C%5C%5C%5C%5Ctext%7BRate%7D_2%3D%5Cfrac%7B2%5Ctimes%202%5E2%7D%7B3%5E%7B1%2F2%7D%7D%5Ctimes%20%28k%5Cfrac%7B%5BA%5D%5BC%5D%5E2%7D%7B%5BB%5D%5E%7B1%2F2%7D%7D%29%5C%5C%5C%5C%5Ctext%7BRate%7D_2%3D4.62%5Ctimes%20%28%5Ctext%7BRate%7D_1%29%5C%5C%5C%5C%5Ctext%7BRate%7D_2%3D4.62%5Ctimes%201.12%5Ctimes%2010%5E%7B-2%7D%5C%5C%5C%5C%5Ctext%7BRate%7D_2%3D5.17%5Ctimes%2010%5E%7B-2%7DM%2Fs)
Hence, the rate of the reaction when concentrations are changed is 
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.