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3 years ago

Write the neutralization equation that would take place in the stomach with two different bases used in antacid products

1 answer:

5 0

In order for the human being stomach to digest food, it uses hydrochloric acid (HCl). This is a very effective an dstrong acid that can burn the lining of the stomach and even cause ulcer if this lining does not contain enough mucus.

Back on track, a reaction between an acid and a base usually yields a salt and water. The two bases that we will use in the reaction with HCl are Mg(OH)2 and CaCO3 (note that (OH)- and (CO3)- both define basis)
MAKE SURE TO BALANCE YOU EQUATION, WHERE THE NUMBER OF MOLECULES ENTERING SHOULD BE EQUAL TO THE NUMBER OF MOLECULES RESULTING.

Equation #1:
2HCl + Mg(OH)2 --> 2H2O + MgCl2
MgCl2 is the salt
Equation#2:
2HCl + CaCO3 --> H2O + CO2 + CaCl2
CaCl2 is the salt
Carbon dioxide is formed as a result of the additional carbon carbon, the reaction ideally outputs H2CO3 which is a very weak base the breaks into water and CO2

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3 years ago

Find the volume. * 1) 6 in 11 in Volume =

posledela

Known :

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Asked :

Volume = ...?

Answer :

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3 years ago

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4 years ago

Read 2 more answers

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

stiv31 [10]

Answer:

Follows are the solution:

Explanation:

A + B = C

Its response decreases over time as well as consumption of a reactants.

r = -kAB

during response A convert into 2x while B convert into x to form 3x of C

let's y = C

y = 3x

Still not converted sum of reaction

for A: 100 - 2x

for B: 50 - x

Shift of x over time

$\frac{dx}{dt} = \frac{-k(100 - 2x)}{(50 - x)}$

Integration of x as regards t

$\frac{1}{[(100 - 2x)(50 - x)]} dx = -k dt\\\\\frac{1}{2[(50 - x)(50 - x)]} dx = -k dt\\\\\ integral\ \frac{1}{2[(50 - x)^2]} dx =\ integral [-k ] \ dt\\\\\frac{-1}{[100-2x]} = -kt + D \\\\$

D is the constant of integration

initial conditions: t = 0, x = 0

$\frac{-1}{[100-2x]} = -kt + D \\\\\frac{ -1}{[100]} = 0 + D\\\\D= \frac{-1}{100}\\\\$

hence we get:

$\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}$

after t = 7 minutes , $C = 10 \ g = 3x$

$3x = 10\\\\x = \frac{10}{3}$

Insert the above value x into $\frac{1}{(100-2x)}$ equation $= kt + \frac{1}{100}$ to get k.

$\to \frac{1}{(100-2\times \frac{10}{3})} = k \times (7) + \frac{1}{100} \\\\ \to \frac{1}{(100- 2 \times 3.33)} = \frac{700k + 1}{100} \\\\ \to \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\$

$\to 100 = 93.34(700k + 1) \\\\ \to 100 = 65,338k + 700 \\\\ \to 65,338k = -600 \\\\ \to k = \frac{-600}{ 65,338} \\\\ \to k= - 0.0091$

therefore plugging in the equation the above value of k

$\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} = \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} = \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}$