Answer:
Moles of magnesium chloride can be produced are 0.2 moles
Explanation:
The reaction of Mg with Cl2 is:
Mg + Cl₂ → MgCl₂
<em>Where 1 mole of Mg reacts per mole of Cl₂ to produce MgCl₂.</em>
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As the reaction is 1:1, we need to convert the mass of both Mg and Cl₂ to moles. The lower number of moles will determine the moles of MgCl₂ that will be produced:
<em>Moles Mg -Molar mass: 24.3g/mol-:</em>
4.86g * (1mol / 24.3g) = 0.2 moles Mg
<em>Moles Cl₂ -Molar mass: 24.3g/mol-:</em>
21.27g * (1mol / 70.9g) = 0.3 moles Cl₂
As moles of Mg < moles of Cl₂, Mg is limiting reactant and moles of magnesium chloride can be produced are 0.2 moles
Compound are formed by two or more elements chemically combined. For example: H^2O is the water formula, this is a compound because you two elements which are Hydrogen and Oxygen and together they form a compound. The (^2) is the amount of atoms the formula has, in this case Hydrogen has two atoms and Oxygen is neutral.
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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+]
⇒ pH = 4.543
