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Over [174]
3 years ago
15

68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi

s reaction, and ____ (compound) is the limiting reagent.
Blank 1:
Blank 2:
Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
7 0

Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

  • 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:

  • 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
  • 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>

Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:

  • 0.528 mol Mg(NO₃)₂ * \frac{1molMg(OH)_2}{1molMg(NO_3)_2} = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

  • 0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g
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Answer:

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2 years ago
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Using the following thermochemical data, what is the change in enthalpy for the following reaction? Ca(OH)2(aq) + HCl(aq) CaCl2(
sesenic [268]
Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?

CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l),  Δ<span>H = -186 kJ
</span>
CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
</span>
1) Ca(OH)2 should be  reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s) 
we are going to take as 
 Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ

2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l),                    and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>

</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)

Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add  enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ
4 0
3 years ago
Two elements that have the same ground-state valence shell configuration of ns 2 np 2 are
kvv77 [185]

Answer:

carbon and silicon

Explanation:

Various groups of elements in the periodic table have different outermost shell electron configurations. Actually, elements are classified into groups on the basis of the number of electrons on the outermost shell of those elements. All elements with the same number of electrons on their outermost shell belong to the same group in the periodic table.

For elements in group 14, they all have four electrons on their outermost shell. Their general outer electron configuration is ns2 np2 as shown in the question. Two prominent members of this group are carbon and silicon. This ns2 np2 is the ground state outer electron configuration of all group 14 elements in the periodic table.

6 0
3 years ago
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vlada-n [284]

Answer:

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Explanation:

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As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.

If we add NaCl to this solution, then we have Na^+ and Cl^- in the solution which will be formed by the ionization of NaCl.

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Due to this reaction the concentration of Zn^2^+ will decrease in the solution and more ZnSe can be soluble in the solution.

6 0
3 years ago
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