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3 years ago

6

Which is the correct formula for the compound made when aqueous solutions containing a dissolved magnesium compound and a dissol

ved chloride compound are mixed?

2 answers:

Gelneren [198K]3 years ago

7 0

Answer:

The correct formula for the compound made when aqueous solutions containing a dissolved Magnesium and Chloride are mixed is Magnesium chloride $MgCl_2$

Explanation:

If we suppose the existence of two components in aqueous solutions containing Magnesium and Chloride. For example: Hydrochloric acid - HCl and Magnesium hydroxide - $Mg(OH)_2$ .

The global reaction would be:

HCl + $Mg(OH)_2$ --> $MgCl_2$ + $H_2 O$

It means, Hydrochloric acid is neutralized with Magnesium hydroxide to produce Magnesium chloride and Water.

Besides of that, we can analyze the aqueos solution of every componenent:

For Hydrochloric acid:

HCl + $H_2 O$ --> $H_3O^+$ + $Cl^-$

For Magnesium hydroxide

$Mg(OH)_2$ + $H_2 O$ --> $H_2O-(OH)^-$ + $Mg^+2$

Finally, the ionic compounds will form the salt:

$Mg^+2$ + $Cl^-$ -> $MgCl_2$

jenyasd209 [6]3 years ago

4 0

It would be MgCl₂ ~ Magnesium Chloride.

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Sixty liters of a gas were collected over water when the barometer read 663 mmhg , and the temperature was 20∘c. what volume wou

lesya [120]

First, let's compute the number of moles in the system assuming ideal gas behavior.

PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
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3 years ago

A 100. ml portion of 0.250 m calcium nitrate solution is mixed with 400. ml of 0.100 m nitric acid solution. what is the final c

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3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

2 years ago

Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2

baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) $Cu_{(s)}$ → $Cu^{2+}_{(aq)} +2e$ E°=0.337 v

(reduction) $MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}$ → $MnO_{2 (aq)}+2H_{2}O$ E°=1.679 v

(overall) $2MnO_{4 (aq)}+3Cu_{(s)}$ +8H^{+}_{(aq)}→ $3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O$ E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

$E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}$

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

$E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346$

E=1.346

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2 years ago