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2 years ago

When excess potassium hydroxide and H2SeO3 react whats the net ionic reaction for the acid-base rxn

Chemistry

1 answer:

Bad White [126]2 years ago

4 0

The required net ionic equation is; 2H^+(aq) + 2OH^-(aq)-----> 2H2O(l)

The molecular reaction equation is;

H2SeO3(aq) + 2KOH(aq) -----> K2SeO3(aq) + 2H2O(l)

The complete ionic equation is;

2H^+(aq) + SeO3^2-(aq) + 2K^+(aq) + 2OH^-(aq)-----> 2K^+(aq) + SeO3^2-(aq) + 2H2O(l)

Net ionic equation;

2H^+(aq) + 2OH^-(aq)-----> 2H2O(l)

We can clearly see that this is a neutralization reaction hence water is the product of the net ionic equation.

Learn more:brainly.com/question/25150590

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what is Calcite's Solubility, I barely know what solubility mean let alone what it has to do with Calcite.

Ivahew [28]

Calcite can be either dissolved by groundwater or precipitated by groundwater, depending on several factors including the water temperature, pH, and dissolved ion concentrations. Although calcite is fairly insoluble in cold water, acidity can cause dissolution of calcite and release of carbon dioxide gas.

7 0

3 years ago

Using the following equation, 2C2H6 +7O2 -->4CO2 +6H2O, if 2.5g C2H6 react with 170g of O2, how many grams of water will be p

kirill [66]

The mass of water (H₂O) that would be produced is 4.5 g

<h3>Stoichiometry </h3>

From the question, we are to determine the mass of water that would be produced.

From the given balanced chemical equation

2C₂H₆ +7O₂ → 4CO₂ +6H₂O

This means

2 moles of C₂H₆ reacts with 7 moles of O₂ to produce 4 moles of CO₂ and 6 moles of H₂O

Now, we will determine the number of moles of each reactant present

For Ethane (C₂H₆)

Mass = 2.5 g

Molar mass = 30.07 g

Using the formula,

$Number\ of\ moles = \frac{Mass}{Molar\ mass}$

Number of moles of C₂H₆ present = $\frac{2.5}{30.07}$

Number of moles of C₂H₆ present = 0.08314 mole

For Oxygen (O₂)

Mass = 170g

Molar mass = 31.999 g/mol

Number of moles of O₂ present = $\frac{170}{31.999}$

Number of moles of O₂ present = 5.3127 moles

Since

2 moles of C₂H₆ reacts with 7 moles of O₂

Then,

0.08314 mole of C₂H₆ will react with $\frac{7 \times 0.08314 }{2}$

$\frac{7 \times 0.08314 }{2}$ = 0.58198 mole

Therefore,

0.08314 mole of C₂H₆ reacts with 0.58198 mole of O₂ to produce 3 × 0.08314 moles of H₂O

3 × 0.08314 = 0.24942 mole

Thus, the number of moles of water (H₂O) produced is 0.24942 mole

Now, for the mass of water that would be produced,

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of water = 18.015 g/mol

Then,

Mass of water that would be produced = 0.24942 × 18.015

Mass of water that would be produced = 4.4933 g

Mass of water that would be produced ≅ 4.5 g

Hence, the mass of water (H₂O) that would be produced is 4.5 g

Learn more on Stoichiometry here: brainly.com/question/14271082

3 0

2 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb ( NO 3 ) 2

nikklg [1K]

Answer : The volume of $NH_4I$ solution required is, 2.93 L

The number of moles of $PbI_2$ formed from the reaction is, 0.662 moles.

Explanation :

First we have to calculate the initial moles of $Pb(NO_3)_2$ .

$\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}$

$\text{Moles of }Pb(NO_3)_2=0.700M\times 0.945L=0.662mol$

Now we have to calculate the moles of $NH_4I$

The balanced chemical reaction is:

$Pb(NO_3)_2(aq)+2NH_4I(aq)\rightarrow PbI_2(s)+2NH_4NO_3(aq)$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react with 2 moles of $NH_4I$

So, 0.662 mole of $Pb(NO_3)_2$ react with $0.662\times 2=1.32$ moles of $NH_4I$

Now we have to calculate the volume of $NH_4I$

$\text{Volume of }NH_4I=\frac{\text{Moles of }NH_4I}{\text{Concentration of }NH_4I}$

$\text{Volume of }NH_4I=\frac{1.32mol}{0.450mol/L}=2.93L$

Now we have to calculate the moles of $PbI_2$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react to give 1 moles of $PbI_2$

So, 0.662 mole of $Pb(NO_3)_2$ react to give 0.662 moles of $PbI_2$

Thus, the number of moles of $PbI_2$ formed from the reaction is, 0.662 moles.

7 0

3 years ago

The rate constant k for a certain reaction is measured at two different temperatures:

bogdanovich [222]

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are $K_1$ and $K_2$ at temperature $T_1$ and $T_2$

By using Arrhenius equation at two different two different temperature,

$Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K ;K_1=4.8*10^8;T_2=273+280=553K ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;$