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Paladinen
2 months ago
13

Based on the results of your titration, what volume of the sample would an adult need to consume to reach the recommended daily

amount? 60 mg is the recommened daily amount of vitamin C.
Ascorbic acid Volume of DCPIP in Total Volume of DCPIP

concentration (g/L) final syringe (mL) used (mL)

0.00 2.18 mL 0.18 mL

0.50 3.1 mL 0.6 mL

1.00 4.6 mL 2.2 mL

1.50 5.56 mL 3.39 mL

2.00 6.37 mL 4.37 mL

Unknown Trial 1 3.21 mL 1.21 mL

Unknown Trial 2 3.11 mL 1.11 mL

Unknown Trial 3 3.15 mL 1.06 mL

Chemistry
1 answer:
eduard [2.7K]2 months ago
5 0
Based on the titration results, the adult would need to consume 85.7 mL to meet the recommended daily intake of 60 mg of Vitamin C. This is calculated from the average total volume of DCPIP used during trials.
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A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium c
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Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

O sea,

Moles =Molarity \times {Volume\ of\ the\ solution}

Dado que:

Para K_2CO_3 :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de K_2CO_3 son:

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles de K_2CO_3 = 0.004 moles

Para Ba(NO_3)_2 :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de Ba(NO_3)_2 son:

Moles=0.400 \times {30.0\times 10^{-3}}\ moles

Moles de Ba(NO_3)_2 = 0.012 moles

Según la reacción:

Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3

1 mol de Ba(NO_3)_2 reacciona con 1 mol de K_2CO_3

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0.012 mol de Ba(NO_3)_2 reacciona con 0.012 mol de K_2CO_3

Moles disponibles de K_2CO_3 = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces K_2CO_3 es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de K_2CO_3 reacciona con 1 mol de Ba(NO_3)_2 y produce 1 mol de BaCO_3

0.004 mol de K_2CO_3 reacciona con 0.004 mol de Ba(NO_3)_2 y genera 0.004 mol de BaCO_3

Los moles restantes de Ba(NO_3)_2 son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, Ba^{2+}, después de la reacción es:

Molarity=\frac{0.008}{0.050}\ M = 0.16\ M

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