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Marianna [84]
3 years ago
11

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

(II) sulfate and aqueous potassium nitrate.
Chemistry
2 answers:
sweet-ann [11.9K]3 years ago
5 0

Explanation: Reaction when aqueous lead (II) nitrate reacts with aqueous potassium sulfate to produce solid lead(II) sulfate and aqueous potassium nitrate is:

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

Ionic equation for the above chemical reaction is:

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

Solid PbSO_4 does not break down because it does not interact with the water molecules.

Net ionic equation for the above reaction is:

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

dedylja [7]3 years ago
3 0

Balanced chemical reaction:  

Pb(NO₃)₂(aq) + K₂SO₄(aq) → PbSO₄(s) + 2KNO₃(aq).

Ionic reaction:

Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + SO₄²⁻(aq) → PbSO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq).

Net ionic reaction:

Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s).

(s) means solid (not dissolved in water).

According to principle of mass conservation, number of atoms must be equal on both side of balanced chemical reaction.  

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What is the energy of a single photon of blue light with a wavelength of 488 nm (488 × 10-9 m)?
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4 years ago
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3 years ago
. How much energy is lost by a 30.0g sample of water that decreases in temperature from 56.7C to 25.0C?
lbvjy [14]

Answer:

Q = -3980.9 j

Explanation:

Given data:

Mass of sample = 30 g

Initial temperature = 56.7 °C

Final temperature = 25 °C

Specific heat of water = 4.186 j/g.°C

Amount of heat released = ?

Formula:

Q = m.c.ΔT

Q = heat released

m = mass of sample

c = specific heat of given sample

ΔT = change in temperature

Solution:

ΔT = T2 -T1

ΔT = 25 °C - 56.7 °C = - 31.7°C

Q = m.c.ΔT

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