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Step2247 [10]
3 years ago
12

Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical equation: ()Zn

+ ()HCl → ()ZnCl2 + ()H2
Chemistry
2 answers:
mote1985 [20]3 years ago
8 0

Answer: 1Zn+2HCl\rightarrow 1ZnCl_2+1H_2

Explanation: A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution.Thus zinc can easily lose electrons as compared to hydrogen and result in the formation of zinc chloride and hydrogen.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

1Zn+2HCl\rightarrow 1ZnCl_2+1H_2

Gekata [30.6K]3 years ago
8 0

Answer:

1,2,1,1

Explanation:

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A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide
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Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

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In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

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n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol

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W=P(V_2-V_1)

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

Thereby, the magnitude and direction of work turn out:

W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

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Regards.

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2 years ago
Calculate the molar solubility and the solubility in g / l of agi at 25°c. the ksp of agi is 8.3 × 10−17.
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Answer is: solubility of silver iodide is 9.11·10⁻⁹ M.<span>.
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