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Aleonysh [2.5K]
3 years ago
12

I need help with #4... Chemistry is my worst subject.

Chemistry
1 answer:
pishuonlain [190]3 years ago
8 0
Hope this helps!! And maybe makes you smile!! I struggle with chemistry as well (and I’m in AP chem ;) )

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Its either A or B, 50-50 chance?
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For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
Nikolay [14]

Explanation:

Molar mass of KClO_{3} = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of O_{2} = 32.0 g

According to the equation, 2 moles of KClO_{3} reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of KClO_{3} will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of KClO_{3} gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of KClO_{3} as follows.

       \frac {96g}{245.2g} \times 2.72 g = 1.06 g of O_{2}


(b) Calculate the amount of oxygen given by 0.361 g of KClO_{3} as follows.

    \frac {96g}{245.2g} \times 0.361 g = 0.141 g of O_{2}


c) Calculate the amount of oxygen given by 83.6 kg KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 83.6 g = 32.731 kg of O_{2}

Convert kg into grams as follows.

    \frac{32.731 kg}{1 kg} \times 1000 g = 32731 g of O_{2}


(d) Calculate the amount of oxygen given by 22.5 mg of KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 22.5 mg = 8.79 mg

Convert mg into grams as follows.

  \frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g of O_{2}

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3 years ago
Which regions on the periodic table can adopt positive and negative oxidation numbers?
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If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
  • pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

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