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2 years ago

How many moles of Mg(OH)2(aq) would be required to neutralize 6.0 mol HCl(aq)?

Chemistry

1 answer:

jekas [21]2 years ago

5 0

Answer:

$3.0molMg(OH)_2$

Explanation:

Hello there!

In this case, for these types of acid-base neutralizations, it is crucial to firstly set up the chemical reaction taking place between the acid and the base; in this case HCl and Mg(OH)2 respectively, whose products are obtained by switching around the anions and cations as shown below:

$HCl(aq)+Mg(OH)_2(aq)\rightarrow MgCl_2(aq)+H_2O(l)$

Which must be balanced to accurately predict the mole ratio on the reactants side:

$2HCl(aq)+Mg(OH)_2(aq)\rightarrow MgCl_2(aq)+2H_2O(l)$

Whereas we can see a 2:1 mole ratio of the acid to the base; thus, the moles of Mg(OH) required for the neutralization of 6.0 moles of HCl turn out to be:

$6.0molHCl*\frac{1molMg(OH)_2}{2molHCl} \\\\=3.0molMg(OH)_2$

Best regards!

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Why the ph of glycine increases when 0.1 M NaOH is added dropwise

shtirl [24]

Answer:

The acid-base reaction produces glycine reduction, and hence the increase of glycine pH.

Explanation:

The glycine is an amino acid with the following chemical formula:

NH₂CH₂COOH

The COOH functional group is what gives the acid properties in the molecule.

Hence, when NaOH is added to glycine an acid-base reaction takes place in which COOH reacts with the NaOH added:

NH₂CH₂COOH + OH⁻ ⇄ NH₂CH₂COO⁻ + H₂O

The glycine concentration starts to shift to its ion form (NH₂CH₂COO⁻) because of the reaction with NaOH, that is why the pH glycine increases when NaOH is added.

Therefore, the acid-base reaction produces glycine reduction, and hence the increase of glycine pH.

I hope it helps you!

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2 years ago

A solution is prepared by dissolving 51 g of salt in 340 g of

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Answer:

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2 years ago

At 150°C the decomposition of acetaldehyde CH3CHO to methane is a first order reaction. If the

Crank

The decomposition time : 7.69 min ≈ 7.7 min

<h3>Further explanation</h3>

Given

rate constant : 0.029/min

a concentration of 0.050 mol L to a concentration of 0.040 mol L

Required

the decomposition time

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time

For first-order reaction :

[A]=[Ao]e^(-kt)

ln[A]=-kt+ln(A0)

Input the value :

ln(0.040)=-(0.029)t+ln(0.050)

-3.219 = -0.029t -2.996

-0.223 =-0.029t

t=7.69 minutes

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2 years ago

Coal can be used to generate hydrogen gas (a potential fuel) by thefollowing endothermic reaction.C(s) + H2O (g) <==> CO(g

SpyIntel [72]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$C(s) + H_2O (g)\leftrightharpoons CO(g) + H_2(g)$

Given that reaction is an endothermic reaction.

For the given options:

a)Adding more C

If the concentration of C that is the reactant is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of C takes place. Therefore, the equilibrium will shift in the right direction to wards the formation of hydrogen gas.

b) Adding more $H_2O$

If the concentration of water that is the reactant is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of water takes place. Therefore, the equilibrium will shift in the right direction towards the formation of hydrogen gas.

c) Raising the temperature of the reaction mixture

If the temperature is increased,heat of the equilibrium mixture will also increase so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in heat that is decrease in temperature occurs.

As, this is an endothermic reaction, forward reaction will decrease the temperature. Hence, the equilibrium will shift in the right direction that is towards the formation of hydrogen gas.

d) Increasing the volume of reaction mixture

If the volume of the container is increased, the pressure will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where increase in pressure is taking place. As the number of moles of gas molecules is greater at the product side. So, the equilibrium will shift in the right direction that is towards the formation of hydrogen gas.

e) Adding a catalyst to reaction mixture

Role of catalyst is to attain the equilibrium quickly without disturbing the state of equilibrium. Hence, addition of catalyst will not change the equilibrium of the reaction.

f) Adding an inert gas to reaction mixture

Adding inert gas to the mixture at constant volume will not effect the equilibrium. Hence, addition of an inert gas will not change the equilibrium of the reaction.

6 0

3 years ago

For each row in the table below, decide whether the pair of elements will form a molecular or ionic compound. If they will, then

Olin [163]

Answer:

$\begin{array}{cccll}\textbf{Element 1} & \textbf{ Element 2} &\textbf{Compound?} &\textbf{Formula} &\textbf{Type}\\\text{Ar}&\text{Xe} &\text{No} &\text{None}&\text{Neither}\\\text{F}& \text{Cs} &\text{Yes} &\text{CsF} &\text{Ionic}\\\text{N} &\text{Br} &\text{Yes} & \text{NBr}_{3}&\text{molecular} \\\end{array}$

Explanation:

You look at the type of atom and their electronegativity difference.

If ΔEN <1.6, covalent; if ΔEN >1.6, ionic

Ar/Xe: Noble gases; no reaction

F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic

N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.

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3 years ago