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3 years ago

Calcium chloride reacts with sodium hydroxide to form solid calcium hydroxide, Ca(OH)2. The balanced net ionic equation is

Chemistry

1 answer:

katen-ka-za [31]3 years ago

5 0

Answer:

Ca²⁺ + 2 OH⁻ → Ca(OH)₂(s)

Explanation:

In chemistry, the net ionic equation is a way to write a chemical reaction whereas you write only the ions that are involved in the reaction.

When calcium chloride, CaCl₂ reacts with sodium hydroxide, NaOH to produce Ca(OH)₂ the only ions involved in the reaction are Ca²⁺ and OH⁻, thus, the balanced net ionic equation is:

Ca²⁺ + 2 OH⁻ → Ca(OH)₂(s)

Cl⁻ and Na⁺ are not involved in the reaction and you don't have to write them.

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Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid. MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l)

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Answer:

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Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.

MnO₂(s) + HCl(aq) → MnCl₂(aq) + H₂O(l) + Cl₂(g)

According to the above reaction, determine the limiting reactant when 5.6 moles of MnO₂ are reacted with 7.5 moles of HCl.

The answer to the above question is

The limiting reactant is the MnO₂

Explanation:

To solve this, we note that one mole of MnO₂ reacts with one mole of HCl to produce one mole of MnCl₂, one mole of H₂O and one mole of Cl₂

Molar mass of MnO₂ = 86.9368 g/mol

Molar mass of HCl = 36.46 g/mol

From the stoichiometry of the reaction, 5.6 moles of MnO₂ will react with 5.6 moles of HCl to produce 5.6 moles of H₂O and 5.6 moles of Cl₂

However there are 7.5 moles of HCL therefore there will be an extra 7.5-5.6 or 1.9 moles of HCl remaining when the reaction is completed

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3 years ago

Can an Atom be broken down into a smaller substance? Yes or No Explain.

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3 years ago

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The normal boiling point of liquid ethyl acetate is 350 K. Assuming that its molar heat of vaporization is constant at 34.4 kJ/m

Roman55 [17]

Answer:

337.22 K

Explanation:

Given that:

P₁ = 1 atm

T₁ = 350 K

P₂ = 0.639 atm

T₂ = ??? (unknown)

R(rate constant) = 8.34 J k⁻¹ mol⁻¹

Using Clausius-Clapeyron equation, we can determine the final boiling point of the process.

Clausius-Clapeyron equation can be written as:

$In\frac{P_2}{P_1}=\frac{\delta H_{vap}}{R}[\frac{T_2-T_1}{T_2T_1}]$

Substituting our values given; we have:

$In\frac{0.639}{1}=(\frac{34.4*10^3J/mol}{8.314 J K^{-1}mol^{-1}})[\frac{T_2-350}{350T_2}]$

$In({0.639})=(\frac{34.4*10^3}{8.314K^{-1}})[\frac{T_2-350}{350T_2}]$

$- 0.4479 = 41317.599 [\frac{T_2-350}{350T_2} ]K$

$-\frac{0.4479}{4137.599} = [\frac{T_2-350}{350T_2} ]$

$- 1.0825118*10^{-4} = [\frac{T_2-350}{350T_2} ]$

$- 1.0825118*10^{-4} *350T_2 =T_2-350$

$- 0.037889T_2=T_2-350$

$350 = 0.03789T_2+T_2$

$350 =1.03789T_2$

$T_2= \frac{350}{1.03789}$

$T_2 = 337.22K$

∴ the boiling point of CH3COOC2H5 when the external pressure is 0.639 atm is 337.22 K.

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3 years ago