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2 years ago

2. The approximate concentration of hydrochloric acid, HCl, in the stomach (stomach

Chemistry

1 answer:

Sergio039 [100]2 years ago

5 0

Answer:

a) 0.714g of bicarbonate of soda are required.

b) 0.221g of Al(OH)₃ are required

Explanation:

The reactions of HCl with bicarbonate of soda and aluminium hydroxide are:

HCl + NaHCO₃ → H₂O + NaCl + CO₂

3 HCl + Al(OH)₃ → 3H₂O + AlCl₃

The moles of HCl that we need neutralize are:

50mL = 0.050L * (0.17mol / L) = 0.0085 moles HCl

To solve these problem we need to find the moles of the antacid using the chemical reaction and its mass using its molar mass;

<em>a) </em><em>Moles NaHCO₃ = Moles HCl = 0.0085 moles </em>

The mass is -Molar mass NaHCO₃: -84g/mol-

0.0085 moles * (84g / mol) = 0.714g of bicarbonate of soda are required

b) 0.0085 moles HCl * (1mol Al(OH)₃ / 3mol HCl) = 2.83x10⁻³ moles Al(OH)₃

The mass is -Molar mass: 78g/mol-:

2.83x10⁻³ moles Al(OH)₃ * (78g/mol) =

<h3>0.221g of Al(OH)₃ are required</h3>

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A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A

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Answer:

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Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

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% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

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3 years ago

Given the partial equation: NO3− Pb2 → NO2 Pb4 , balance the reaction in acidic solution using the half-reaction method and fill

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Answer : The balanced reaction in acidic solution is,

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$NO_3^-+Pb^{2+}\rightarrow NO_2+Pb^{4+}$

First we have to separate into half reaction. The two half reactions are:

$NO_3^-\rightarrow NO_2$

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Now we have to balance the half reactions in acidic medium, we get:

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$Pb^{2+}\rightarrow Pb^{4+}+2e^-$ ............(2)

Now we have to balance the electrons of the half reactions. When we are multiplying the equation (1) by 2, we get

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Now we have to add both the half reactions (2) and (3), we get the final balanced chemical reaction.

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3 years ago