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KATRIN_1 [288]
2 years ago
11

PLSSS HELP MEECH3COOH + MgCH3-COOH+MgOCH3 -COOH + Mg(OH)2CH3-COOH + MgCO3​

Chemistry
1 answer:
iogann1982 [59]2 years ago
4 0

Explanation:

CH3COOH + Mg --> Mg(CH3COO)2 + H2

CH3COOH + MgO --> H2O + Mg(CH3COO)2

CH3COOH + Mg(OH)2 --> H2O + (CH3COO)2Mg

CH3COOH + MgCO3 --> (CH3COO)2Mg + CO2 + H2O.

<em>*</em><em>equations</em><em> </em><em>are </em><em>not </em><em>balanced.</em>

hope it helps. :)

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Amino acids can be synthesized by reductive amination. Draw the structure of the organic compound that you would use to synthesi
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Answer:

Following are the answer to this question:

Explanation:

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3 years ago
What are the products in this chemical reaction? 2H3PO4+3Ca(OH)2→Ca3(PO4)2+6H2O
finlep [7]
<h2>Answer:C</h2>

Explanation:

A chemical equation is a equation that describes a corresponding chemical reaction.

A chemical reaction is generally written as

reactants→products

reactants refer to all the reactants involved in the chemical reaction.

Reactants are usually written on the left hand side of the chemical equation.

products refer to all the products formed in the chemical reaction.

products are usually written on the right hand side of the chemical equation.

In the given reaction,Ca_{3}(PO_{4})_{2},H_{2}O are written on the right side of the equation.

So,Ca_{3}(PO_{4})_{2},H_{2}O are the products.

5 0
2 years ago
Read 2 more answers
The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−
barxatty [35]

Answer:

The reaction will be  non spontaneous at these concentrations.

Explanation:

AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)

Expression for an equilibrium constant K_c:

K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]

Solubility product of the reaction:

K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}

Reaction between Gibb's free energy and equilibrium constant if given as:

\Delta G^o=-2.303\times R\times T\times \log K_c

\Delta G^o=-2.303\times R\times T\times \log K_{sp}

\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]

\Delta G^o=69,117.84 J/mol=69.117 kJ/mol

Gibb's free energy when concentration [Ag^+] = 1.0\times 10^{-2} M and [Br^-] = 1.0\times 10^{-3} M

Reaction quotient of an equilibrium = Q

Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}

\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)

\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])

\Delta G=40.588 kJ/mol

  • For reaction to spontaneous reaction:  \Delta G.
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Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0
3 years ago
Find the empirical formula of the following compounds:
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The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

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