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2 years ago

How can I balance this equation? KClO3 ---> KCl + O2

1 answer:

8 0

__ KClO₃ → __ KCl + __ O₂

Left Side:
1 K
1 Cl
3 O

Right Side:
1 K
1 Cl
2 O

Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.

This gives us 2KClO₃ → __ KCl + 3O₂.

However, this equation is still not balanced.

Left Side:
2 K
2 Cl
6 O

Right Side:
1 K
1 Cl
6 O

In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.

2KClO₃ → 2KCl + 3O₂

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A sodium hydroxide solution is made by mixing 8.70 g NaOH with 100 g of water. The resulting solution has a density of 1.087 g/m

Ahat [919]

Answer:

Mass fraction = 0.08004

Mole fraction = 0.0377

Explanation:

Given, Mass of NaOH = 8.70 g

Mass of solution = 8.70 + 100 g = 108.70 g

$Mass\ fraction\ of\ NaOH=\frac {8.70}{108.70}$ = 0.08004

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{8.70\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 0.2175\ mol$

Given, Mass of water = 100 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{100\ g}{18.0153\ g/mol}$

$Moles\ of\ water= 5.5508\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ NaOH=\frac {n_{NaOH}}{n_{NaOH}+n_{water}}$

$Mole\ fraction\ of\ NaOH=\frac {0.2175}{0.2175+5.5508}=0.0377$

3 0

2 years ago

When the following aqueous solutions are mixed together, a precipitate forms. Balance the net ionic equation in standard form fo

rewona [7]

Answer:

For (a): The balanced net ionic equation is $2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$ and the sum of coefficients is 4

For (b): The balanced net ionic equation is $Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$ and the sum of coefficients is 4

For (c): The balanced net ionic equation is $Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$ and the sum of coefficients is

For (d): The balanced net ionic equation is $Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$ and the sum of coefficients is 4

For (e): The balanced net ionic equation is $Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ and the sum of coefficients is 3

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

$Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

$2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

$K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)$

The complete ionic equation follows:

$2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

$2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

$Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

$NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)$

The complete ionic equation follows:

$Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

8 0

3 years ago

A compound has the empirical formula . a 256-ml flask, at 373 k and 750. torr, contains 0.527 g of the gaseous compound. give th

CaHeK987 [17]

Answer: From the ideal gas law, MM=mRTPV; where MM = molecular mass; m = mass; P = pressure in atmospheres; V= volume in litres; R = gas constant with appropriate units. So, 0.800â‹…gĂ—0.0821â‹…Lâ‹…atmâ‹…Kâ’1â‹…molâ’1Ă—373â‹…K0.256â‹…LĂ—0.987â‹…atm = 97.0 gâ‹…molâ’1. nĂ—(12.01+1.01+2Ă—35.45)â‹…gâ‹…molâ’1 = 97.0â‹…gâ‹…molâ’1. Clearly, n = 1. And molecular formula = C2H2Cl2. I seem to recall (but can't be bothered to look up) that vinylidene chloride, H2C=C(Cl)2 is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required.

4 0

3 years ago

A first order reaction has a rate constant of 0.543 at 25 c and 6.47 at 47

alukav5142 [94]

The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

where :

k1 is the rate constant at temperature T1

k2 is the rate constant at temperature T2

R = gas constant = 8.314 J/K-mol

Given data:

k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K

k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K

ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]

2.478 = 2.774 *10^-5 Ea

Ea = 0.8934*10^5 J = 89.3 kJ

5 0

2 years ago

For all of the following questions 20.00 mL of 0.200 M HBr is titrated with 0.200 M KOH.

HACTEHA [7]

Answer :

The concentration of $H^+$ before any titrant added to our starting material is 0.200 M.

The pH based on this $H^+$ ion concentration is 0.698

Explanation :

First we have to calculate the concentration of $H^+$ before any titrant is added to our starting material.

As we are given:

Concentration of HBr = 0.200 M

As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion $H^+$ and bromide ion $Br^-$ .

As, 1 M of HBr dissociates to give 1 M of $H^+$

So, 0.200 M of HBr dissociates to give 0.200 M of $H^+$

Thus, the concentration of $H^+$ before any titrant added to our starting material is 0.200 M.

Now we have to calculate the pH based on this $H^+$ ion concentration.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (0.200)$

$pH=0.698$

Thus, the pH based on this $H^+$ ion concentration is 0.698

3 0

3 years ago

How can I balance this equation? ____ KClO3 ---> ____ KCl + ____ O2

How can I balance this equation? KClO3 ---> KCl + O2