A ___________________________ is a mixture in which the properties of thesubstances are spread evenly.

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in basic aqueous solution. Be sure to add p

Butoxors [25]

Answer:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Explanation:

Hello there!

In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

$BiO_3^-\rightarrow Bi^{3+}$

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

$(Bi^{5+}O_3)^-\rightarrow Bi^{3+}$

Thereafter, we realize three water molecules are needed on the right in order to balance the oxygens and consequently 6 hydrogen atoms on the left to balance hydrogen:

$6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O$

Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

$6OH^-+6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O-3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Then, we accommodate the waters to obtain:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Best regards!

7 0

3 years ago

Identify the correct net ionic equation for the reaction that occurs when solutions of Pb(NO3)2 and NH4Cl are mixed.

Pie

Answer:

Pb2+(aq) + 2Cl–(aq) ----> PbCl2(s)

Explanation:

The net ionic equation shows the main reaction that takes place in a system. Hence, a net ionic equation focusses only on those species that actually participate in the reaction.

For the reaction between Pb(NO3)2 and NH4Cl , the net ionic equation is;

Pb^+(aq) + 2Cl^-(aq) ---> PbCl2(s)

7 0

3 years ago

What is the volume occupied by 4.20 miles of oxygen gas (O2) at STP

Drupady [299]

Answer: 94.13 L

Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.

To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.

8 0

2 years ago

A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas

sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4 × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$ $\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$