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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:
<span>2HCl + Mg(OH)2 --> 2H2O + MgCl2
MgCl2 is the salt
</span>Equation#2:
<span>2HCl + CaCO3 --> H2O + CO2 + CaCl2</span>
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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Write a conclusion using the CER method. Your conclusion must include the following: - Claim: State your claim. Your claim is th

Karolina [17]

Nobody on here is going to write a entire cer for you

4 0

3 years ago

What is the density of an object that has a mass of 6.5 g and when placed in water displaces the volume from 4.5mL to 11.8mL? ro

KengaRu [80]

Answer:

$\rho =0.9g/mL$

Explanation:

Hello,

In this case, due to the volume displacement caused the by the object's submersion, it's volume is:

$V=11.8mL-4.5mL=7.3 mL$

In such a way, considering the mathematical definition of density, it turns out:

$\rho =\frac{m}{V}=\frac{6.5g}{7.3mL}\\ \\\rho =0.89g/mL$

Rounding to the nearest tenth we finally obtain:

$\rho =0.9g/mL$

Regards.

3 0

3 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

3 years ago

Write empirical formula

andrew11 [14]

Answer:

$Pb(ClO_{3})_{4}\\Pb(MnO_{4})_{4}\\Fe(ClO_{3})_{3}\\\Fe(MnO_{4})_{3}\\$

Explanation:

$Pb^{4+}(ClO_{3}^{-})_{4}--->Pb(ClO_{3})_{4}\\Pb^{4+}(MnO_{4}^{-})_{4}--->Pb(MnO_{4})_{4}\\Fe^{3+}(ClO_{3}^{-})_{3}--->Fe(ClO_{3})_{3}\\\Fe^{3+}(MnO_{4}^{-})_{3}--->Fe(MnO_{4})_{3}\\$

3 0

3 years ago

Which atom would be expected to have a half-filled 6s subshell

eduard

Answer: I think the answer is Cesium (Cs)

Explanation:

A half-filled 6s subshell would be 6s^1

4 0

3 years ago