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3 years ago

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net ionic equation to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II) sulfat

e and aqueous potassium nitrate.

2 answers:

vladimir1956 [14]3 years ago

4 0

<u>Answer:</u> the net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of lead (II) nitrate and potassium sulfate is given as:

$Pb(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow PbSO_4(s))+2KNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2K^+(aq.)+SO_4^{2-}(aq.)\rightarrow PbSO_4(s)+2K^+(aq.)+2NO_3^-(aq.)$

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow PbSO_4(s)$

Hence, the net ionic equation is written above.

Masja [62]3 years ago

3 0

Net ionic equation: Pb²⁺ + SO₄²⁻ → PbSO₄

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Balance the following reaction.<br> A.) 4<br> B.) 2<br> C.) 5<br> D.) blank<br> E.) 3

slega [8]

Answer:

C.5

Hope this helps

7 0

2 years ago

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

4 0

3 years ago

What element is oxidized in the chemical reaction 2Mn(NO3)2 + 5NaBiO3 + 16HNO3 → 2HMnO4 + 5Bi(NO3)3 + 5NaNO3 + 7H2O

Lena [83]

I’m not sure lol good luck

5 0

2 years ago

!!!Need answer for chem homework ASAP PLS !!!!

stealth61 [152]

Answer:

The answer to your question is: kc = 6.48

Explanation:

Data

Given Molecular weight

CaO = 44.6 g 56 g

CO₂ = 26 g 44 g

CaCO₃ = 42.3 g 100 g

Find moles

CaO 56 g ---------------- 1 mol

44.6 g -------------- x

x = (44.6 x 1) / 56 = 0.8 mol

CO₂ 44 g ----------------- 1 mol

26 g ---------------- x

x = (26 x 1 ) / 44 = 0.6 moles

CaCO₃ 100 g --------------- 1 mol

42.3g -------------- x

x = (42.3 x 1) / 100 = 0.423 moles

Concentrations

CaO = 0.8 / 6.5 = 0.12 M

CO₂ = 0.6 / 6.5 = 0.09 M

CaCO₃ = 0.423 / 6.5 = 0.07 M

Equilibrium constant = $\frac{[products]}{[reactants]}$

Kc = [0.07] / [[0.12][0.09]

Kc = 0.07 / 0.0108

kc = 6.48

7 0

3 years ago

A sample of N2(g) was collected over water at 25°C and 730 torr in a container with a volume of 340 mL. The vapor pressure of wa

AnnZ [28]

Answer:

D. 0.36 g

Explanation:

When a gas is collected over water, the total pressure is the sum of the pressure of the gas and the pressure of the water vapor.

Ptotal = Pwater + PN₂

PN₂ = Ptotal - Pwater = 730 torr - 23.76 torr = 706 torr

We can find the mass of N₂ using the ideal gas equation.

$P.V=n.R.T=\frac{m}{M} .R.T\\m=\frac{P.V.M}{R.T} =\frac{730torr.0.340L.28g/mol}{(0.08206atm.L/mol.K).298K} .\frac{1atm}{760torr} =0.36g$

6 0

2 years ago