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zavuch27 [327]
3 years ago
13

For the elementary reaction NO3 + CO → NO2 + CO2 the molecularity of the reaction is ________, and the rate law is rate = ______

__. For the elementary reaction NO3 + CO NO2 + CO2 the molecularity of the reaction is ________, and the rate law is rate = ________. 2, k[NO3][CO]/[NO2][CO2] 4, k[NO3][CO][NO2][CO2] 2, k[NO2][CO2] 4, k[NO2][CO2]/[NO3][CO] 2, k[NO3][CO]
Chemistry
1 answer:
blagie [28]3 years ago
3 0

Answer: For the elementary reaction NO_3+CO\rightarrow NO_2+CO_2 the molecularity of the reaction is 2, and the rate law is rate = k[NO_3]^1[CO]^1

Explanation:

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.  Thus it can never be fractional.

For elementary reaction NO_3+CO\rightarrow NO_2+CO_2 , molecularity is 2 and rate law is rate=k[NO_3]^1[CO]^1

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In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 gram
kherson [118]

This question is incomplete, the complete question is;

In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization

Compound      Solubility in water at 25°C      Solubility in water at 100°C

Benzoic Acid      0.34 g/100mL                           5.6 g/100mL

Acetanilide         0.53 g/100mL                           5.5 g/100mL

Answer:

The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL

Explanation:

Given the data in the question;

mass of sample = 5 g

percentage composition of the acetanilide impurity = 6.3%

mass of the acetanilide in impure sample will be;

⇒ 6.3% × 5 g = 0.315 g

Mass of benzoic acid in impure sample;

⇒ 5 g - 0.315 g = 4.685 g

now, solubility in water at 100°C for benzoic acid = 5.6 g/100mL

hence 4.685 g of benzoic acid is soluble in x mL

x = [ 100 mL × 4.685 g ] / 5.6 g

x = 83.66 ≈ 84 mL

Also, solubility in water at 100°C for acetanilide = 5.5 g/100mL

hence 0.315 g of benzoic acid is soluble in x mL

x = [ 100 mL × 0.315 g ] / 5.5 g

x = 5.727 ≈ 6 mL

So, the minimum amount in mL of solvent (water) required for the recrystallization will be;

⇒ 85 mL + 6 mL = 91 mL

The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL

5 0
3 years ago
How many grams of iki would it take to obtain a 100 ml solution of 0.300 m iki? how many grams of iki would it take to create a
GalinKa [24]

0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.

By using a conversion factor:

100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g

Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.

In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g

3 0
3 years ago
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